riaology
Calculus II / 102 practice questions

Choose the method.
Make the connection.

Draw the situation, choose a relationship, and work the problem before opening its visual solution. Search for a topic or mix the order to practice choosing the method.

Back to the course ↗

34 lesson question sets found · Page 1 of 6.

Integration techniques

Transfer a derivative to the simpler factor.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper time b 2 . State its sign and units.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Start with the product derivative, then integrate and rearrange. Choose u so its derivative simplifies and dv so its integral is known. Include the boundary term; for definite integrals it can carry much of the result.

Explore the visual solution
Read the model∫0bte−t dt=[−(t+1)e−t]0b=1−(b+1)e−b\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

Input t versus t exp(−t). Integrand. Selected definite integral: 0.59399 . Integration by parts transfers the derivative from one factor to another, using the product rule backward.000.5112t exp(−t)Input t
Integrand

definite integralI=1−(2+1)e−2=0.593994I=1-(2+1)e^{-2}=0.593994

Integration by parts transfers the derivative from one factor to another, using the product rule backward.

Assumptions, units, and limits

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Start with the product derivative, then integrate and rearrange. Choose u so its derivative simplifies and dv so its integral is known. Include the boundary term; for definite integrals it can carry much of the result.

∫0bte−t dt=[−(t+1)e−t]0b=1−(b+1)e−b\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}
Calculate and explain

Apply the idea to a new scenario

Integrate t exp(−t) from 0 to 3.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Start with the product derivative, then integrate and rearrange. Choose u so its derivative simplifies and dv so its integral is known. Include the boundary term; for definite integrals it can carry much of the result.

Explore the visual solution
Read the model∫0bte−t dt=[−(t+1)e−t]0b=1−(b+1)e−b\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

Input t versus t exp(−t). Integrand. Selected definite integral: 0.80085 . Integration by parts transfers the derivative from one factor to another, using the product rule backward.000.51.513t exp(−t)Input t
Integrand

definite integralI=1−(3+1)e−3=0.800852I=1-(3+1)e^{-3}=0.800852

Integration by parts transfers the derivative from one factor to another, using the product rule backward.

Assumptions, units, and limits

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Start with the product derivative, then integrate and rearrange. Choose u so its derivative simplifies and dv so its integral is known. Include the boundary term; for definite integrals it can carry much of the result.

∫0bte−t dt=[−(t+1)e−t]0b=1−(b+1)e−b\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the model∫0bte−t dt=[−(t+1)e−t]0b=1−(b+1)e−b\int_0^bte^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^b=1-(b+1)e^{-b}

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

Input t versus t exp(−t). Integrand. Selected definite integral: 0.59399 . Integration by parts transfers the derivative from one factor to another, using the product rule backward.000.5112t exp(−t)Input t
Integrand

definite integralI=1−(2+1)e−2=0.593994I=1-(2+1)e^{-2}=0.593994

Integration by parts transfers the derivative from one factor to another, using the product rule backward.

Assumptions, units, and limits

Use u=tu=t and dv=edv=e^(−t)dt. Choosing the polynomial for u makes its derivative simpler.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Start with the product derivative, then integrate and rearrange. Choose u so its derivative simplifies and dv so its integral is known. Include the boundary term; for definite integrals it can carry much of the result.

Integration techniques

An identity can expose a simpler integrand.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper angle b 1.57 rad. State its sign and units.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

An even sine power is rewritten with a half-angle identity. With odd powers, reserve one sine or cosine factor and use the remaining squared identity for substitution. The choice depends on the integrand's parity.

Explore the visual solution
Motion illustration: position 0.9999996829318346. Velocity direction shown where applicable.Position 1 · arrow: velocity direction
Angle x (rad) versus sin²(x). Squared oscillation. Selected definite integral: 0.7846 . For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.000.50.78511.57sin²(x)Angle x (rad)
Squared oscillation

definite integralI=1.572−sin⁡(3.14)4=0.784602I=\frac{1.57}2-\frac{\sin(3.14)}{4}=0.784602

For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.

Assumptions, units, and limits

Radian angles; the half-angle identity reduces an even power.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

An even sine power is rewritten with a half-angle identity. With odd powers, reserve one sine or cosine factor and use the remaining squared identity for substitution. The choice depends on the integrand's parity.

sin⁡2x=1−cos⁡2x2,∫0bsin⁡2x dx=b2−sin⁡2b4\sin^2x=\frac{1-\cos2x}{2},\quad\int_0^b\sin^2x\,dx=\frac b2-\frac{\sin2b}{4}
Calculate and explain

Apply the idea to a new scenario

Integrate sin²x over 0 to π/2 radians.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

An even sine power is rewritten with a half-angle identity. With odd powers, reserve one sine or cosine factor and use the remaining squared identity for substitution. The choice depends on the integrand's parity.

Explore the visual solution
Motion illustration: position 1. Velocity direction shown where applicable.Position 1 · arrow: velocity direction
Angle x (rad) versus sin²(x). Squared oscillation. Selected definite integral: 0.7854 . For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.000.50.78511.571sin²(x)Angle x (rad)
Squared oscillation

definite integralI=1.57082−sin⁡(3.14159)4=0.785398I=\frac{1.5708}2-\frac{\sin(3.14159)}{4}=0.785398

For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.

Assumptions, units, and limits

Radian angles; the half-angle identity reduces an even power.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

An even sine power is rewritten with a half-angle identity. With odd powers, reserve one sine or cosine factor and use the remaining squared identity for substitution. The choice depends on the integrand's parity.

sin⁡2x=1−cos⁡2x2,∫0bsin⁡2x dx=b2−sin⁡2b4\sin^2x=\frac{1-\cos2x}{2},\quad\int_0^b\sin^2x\,dx=\frac b2-\frac{\sin2b}{4}
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Motion illustration: position 0.9999996829318346. Velocity direction shown where applicable.Position 1 · arrow: velocity direction
Angle x (rad) versus sin²(x). Squared oscillation. Selected definite integral: 0.7846 . For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.000.50.78511.57sin²(x)Angle x (rad)
Squared oscillation

definite integralI=1.572−sin⁡(3.14)4=0.784602I=\frac{1.57}2-\frac{\sin(3.14)}{4}=0.784602

For odd sine/cosine powers, reserve one factor for substitution. Even powers often need half-angle identities.

Assumptions, units, and limits

Radian angles; the half-angle identity reduces an even power.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. An even sine power is rewritten with a half-angle identity. With odd powers, reserve one sine or cosine factor and use the remaining squared identity for substitution. The choice depends on the integrand's parity.

Integration techniques

Turn a radical into a trigonometric identity.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper x bound b 1 . State its sign and units.

Checked quantity: definite integral · square units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Substitute x=a sinθ and include dx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

Explore the visual solution
Read the modelx=2sin⁡θ,∫0b4−x2 dx=b24−b2+2arcsin⁡(b/2)x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Input x versus sqrt(4−x²). Circle height. Selected definite integral: 1.91322 square units. The substitution removes the radical by an identity; dx=2cosθ dθ also changes. A reference triangle converts back to x.0010.521sqrt(4−x²)Input x
Circle height

definite integralI=124−12+2arcsin⁡(1/2)=1.91322I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322

The substitution removes the radical by an identity; dx=2dx=2cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Substitute x=ax=a sinθ and include dx=adx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

x=2sin⁡θ,∫0b4−x2 dx=b24−b2+2arcsin⁡(b/2)x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)
Calculate and explain

Apply the idea to a new scenario

Integrate sqrt(4−x24-x^2) from 0 to 1.5.

Checked quantity: definite integral · square units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Substitute x=a sinθ and include dx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

Explore the visual solution
Read the modelx=2sin⁡θ,∫0b4−x2 dx=b24−b2+2arcsin⁡(b/2)x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Input x versus sqrt(4−x²). Circle height. Selected definite integral: 2.68828 square units. The substitution removes the radical by an identity; dx=2cosθ dθ also changes. A reference triangle converts back to x.0010.7521.5sqrt(4−x²)Input x
Circle height

definite integralI=1.524−1.52+2arcsin⁡(1.5/2)=2.68828I=\frac{1.5}2\sqrt{4-1.5^2}+2\arcsin(1.5/2)=2.68828

The substitution removes the radical by an identity; dx=2dx=2cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Substitute x=ax=a sinθ and include dx=adx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

x=2sin⁡θ,∫0b4−x2 dx=b24−b2+2arcsin⁡(b/2)x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelx=2sin⁡θ,∫0b4−x2 dx=b24−b2+2arcsin⁡(b/2)x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Input x versus sqrt(4−x²). Circle height. Selected definite integral: 1.91322 square units. The substitution removes the radical by an identity; dx=2cosθ dθ also changes. A reference triangle converts back to x.0010.521sqrt(4−x²)Input x
Circle height

definite integralI=124−12+2arcsin⁡(1/2)=1.91322I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322

The substitution removes the radical by an identity; dx=2dx=2cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Substitute x=ax=a sinθ and include dx=adx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

Integration techniques

A hanging cable links exponentials and arc length.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the cable slope using cable parameter a 1.5 m, horizontal x 1 m. State its sign and units.

Checked quantity: cable slope · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Explore the visual solution
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.71716 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.52103.0432Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

y=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1
Calculate and explain

Apply the idea to a new scenario

For cable parameter a=2a=2 m at x=1x=1 m, find the slope.

Checked quantity: cable slope · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Explore the visual solution
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.5211 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.54303.0862Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/2)=0.521095y'(1)=\sinh(1/2)=0.521095

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

y=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.71716 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.52103.0432Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Integration techniques

Break a rational integrand into recognizable pieces.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper bound b 2 . State its sign and units.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

Explore the visual solution
Read the model1(x+1)(x+2)=1x+1−1x+2,∫0b=ln⁡2(b+1)b+2\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Input x versus 1/((x+1)(x+2)). Rational integrand. Selected definite integral: 0.40547 . Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.000.51121/((x+1)(x+2))Input x
Rational integrand

definite integralI=ln⁡2(2+1)2+2=0.405465I=\ln\frac{2(2+1)}{2+2}=0.405465

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

1(x+1)(x+2)=1x+1−1x+2,∫0b=ln⁡2(b+1)b+2\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}
Calculate and explain

Apply the idea to a new scenario

Integrate 1/[(x+1x+1)(x+2x+2)] from 0 to 4.

Checked quantity: definite integral · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

Explore the visual solution
Read the model1(x+1)(x+2)=1x+1−1x+2,∫0b=ln⁡2(b+1)b+2\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Input x versus 1/((x+1)(x+2)). Rational integrand. Selected definite integral: 0.51083 . Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.000.52141/((x+1)(x+2))Input x
Rational integrand

definite integralI=ln⁡2(4+1)4+2=0.510826I=\ln\frac{2(4+1)}{4+2}=0.510826

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

1(x+1)(x+2)=1x+1−1x+2,∫0b=ln⁡2(b+1)b+2\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the model1(x+1)(x+2)=1x+1−1x+2,∫0b=ln⁡2(b+1)b+2\frac1{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2},\quad\int_0^b=\ln\frac{2(b+1)}{b+2}

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

Input x versus 1/((x+1)(x+2)). Rational integrand. Selected definite integral: 0.40547 . Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.000.51121/((x+1)(x+2))Input x
Rational integrand

definite integralI=ln⁡2(2+1)2+2=0.405465I=\ln\frac{2(2+1)}{2+2}=0.405465

Solve coefficient identities after multiplying by the denominator. The two logarithmic terms combine after evaluating endpoints.

Assumptions, units, and limits

Distinct linear factors, no poles on the integration interval. Partial fractions requires proper rational form first.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. First ensure the rational expression is proper; otherwise perform polynomial division. Distinct linear factors need constant numerators, repeated factors need all powers, and irreducible quadratic factors need linear numerators. Solve a polynomial identity for coefficients.

Integration techniques

An infinite tail needs a limiting argument.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the finite-cutoff integral (not infinite limit) using exponent p 2 , finite upper cutoff b 10 . State its sign and units.

Checked quantity: finite-cutoff integral (not infinite limit) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Replace the infinite endpoint with a finite cutoff, integrate, then take the cutoff limit. A singular interior point needs separate one-sided integrals. Divergent pieces cannot be canceled to claim an ordinary improper integral converges.

Explore the visual solution
Read the model∫1Bx−p dx={B1−p−11−pp≠1ln⁡Bp=1,∫1∞x−p dx converges iff p>1\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

x versus x^(−p). Tail integrand. Selected finite-cutoff integral (not infinite limit): 0.9 . A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.000.55110x^(−p)x
Tail integrand

finite-cutoff integral (not infinite limit)IB=0.9,I∞=12−1=1I_B=0.9,\quad I_\infty=\frac1{2-1}=1

A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.

Assumptions, units, and limits

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Replace the infinite endpoint with a finite cutoff, integrate, then take the cutoff limit. A singular interior point needs separate one-sided integrals. Divergent pieces cannot be canceled to claim an ordinary improper integral converges.

∫1Bx−p dx={B1−p−11−pp≠1ln⁡Bp=1,∫1∞x−p dx converges iff p>1\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1
Calculate and explain

Apply the idea to a new scenario

With exponent p=1p=1 and cutoff B=20,B=20, find finite accumulated area and classify the infinite tail.

Checked quantity: finite-cutoff integral (not infinite limit) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Replace the infinite endpoint with a finite cutoff, integrate, then take the cutoff limit. A singular interior point needs separate one-sided integrals. Divergent pieces cannot be canceled to claim an ordinary improper integral converges.

Explore the visual solution
Read the model∫1Bx−p dx={B1−p−11−pp≠1ln⁡Bp=1,∫1∞x−p dx converges iff p>1\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

x versus x^(−p). Tail integrand. Selected finite-cutoff integral (not infinite limit): 2.99573 . A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.000.510120x^(−p)x
Tail integrand

finite-cutoff integral (not infinite limit)IB=2.99573,I∞ divergesI_B=2.99573,\quad I_\infty\text{ diverges}

A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.

Assumptions, units, and limits

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Replace the infinite endpoint with a finite cutoff, integrate, then take the cutoff limit. A singular interior point needs separate one-sided integrals. Divergent pieces cannot be canceled to claim an ordinary improper integral converges.

∫1Bx−p dx={B1−p−11−pp≠1ln⁡Bp=1,∫1∞x−p dx converges iff p>1\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the model∫1Bx−p dx={B1−p−11−pp≠1ln⁡Bp=1,∫1∞x−p dx converges iff p>1\int_1^Bx^{-p}\,dx=\begin{cases}\frac{B^{1-p}-1}{1-p}&p\ne1\\\ln B&p=1\end{cases},\quad\int_1^\infty x^{-p}\,dx\text{ converges iff }p>1

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

x versus x^(−p). Tail integrand. Selected finite-cutoff integral (not infinite limit): 0.9 . A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.000.55110x^(−p)x
Tail integrand

finite-cutoff integral (not infinite limit)IB=0.9,I∞=12−1=1I_B=0.9,\quad I_\infty=\frac1{2-1}=1

A finite-looking plot never proves convergence. The limiting accumulated area decides whether an infinite tail has finite total.

Assumptions, units, and limits

A finite cutoff is not the improper integral. Convergence requires the limit as B tends to infinity.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Replace the infinite endpoint with a finite cutoff, integrate, then take the cutoff limit. A singular interior point needs separate one-sided integrals. Divergent pieces cannot be canceled to claim an ordinary improper integral converges.