Integrals Practice

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Difficulty

Direct

1.1Use the Power Rule and Linearity

Exam I | Problem 1.1 |

Evaluate the indefinite integral:

\int \left(4x^3 - 6x + 2\right)\,dx

Solution

Integrate term by term:

\int 4x^3\,dx = x^4

\int (-6x)\,dx = -3x^2

\int 2\,dx = 2x

So,

\int \left(4x^3 - 6x + 2\right)\,dx = x^4 - 3x^2 + 2x + C.

1.2Differentiate an Accumulation Function

Exam I | Problem 1.2 |

Let

F(x) = \int_0^x (t^2 + 3t)\,dt.

Find $F'(x)$.

Solution

By Part 1 of the Fundamental Theorem of Calculus,

$$ F'(x) = x^2 + 3x. $$

1.3Evaluate a Definite Integral with the FTC

Exam I | Problem 1.3 |

Compute:

\int_0^2 \left(3x^2 - 4x + 1\right)\,dx

Solution

An antiderivative is

$$ x^3 - 2x^2 + x. $$

Now apply Part 2 of the Fundamental Theorem of Calculus:

\int_0^2 \left(3x^2 - 4x + 1\right)\,dx = \left[x^3 - 2x^2 + x\right]_0^2

$$ = (8 - 8 + 2) - 0 = 2. $$

1.4Find an Average Value

Exam I | Problem 1.4 |

Find the average value of

$$ f(x) = x^2 $$

on the interval $[0,3]$.

Solution

Use the average value formula:

f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\,dx.

Compute the integral:

\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = 9.

So the average value is

\frac{1}{3}\cdot 9 = 3.

1.5Use Substitution on a Composite Function

Exam I | Problem 1.5 |

Evaluate:

\int 2x\cos(x^2)\,dx

Solution

Let

u = x^2, \qquad du = 2x\,dx.

Then the integral becomes

\int \cos u\,du = \sin u + C.

Substitute back:

\int 2x\cos(x^2)\,dx = \sin(x^2) + C.

1.6Use Integration by Parts on a Product

Exam I | Problem 1.6 |

Evaluate:

\int x e^x\,dx

Solution

Use integration by parts with

u = x, \qquad dv = e^x\,dx.

Then

du = dx, \qquad v = e^x.

\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C.

1.7Decompose a Simple Rational Integral

Exam I | Problem 1.7 |

Evaluate:

\int \frac{1}{x(x+2)}\,dx

Solution

Write the integrand as partial fractions:

\frac{1}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2}.

Multiply by $x(x+2)$:

$$ 1 = A(x+2) + Bx. $$

Set $x=0$ to get $1=2A$, so $A=\frac12$. Set $x=-2$ to get $1=-2B$, so $B=-\frac12$.

Thus

\frac{1}{x(x+2)} = \frac{1}{2x} - \frac{1}{2(x+2)}.

Integrate:

\int \frac{1}{x(x+2)}\,dx = \frac12 \ln|x| - \frac12 \ln|x+2| + C.

1.8Use a Trig Identity Before Integrating

Exam I | Problem 1.8 |

Compute:

\int_0^{\pi/2} \sin^2 x\,dx

Solution

Use the identity

\sin^2 x = \frac{1-\cos(2x)}{2}.

Then

\int_0^{\pi/2} \sin^2 x\,dx = \frac12 \int_0^{\pi/2} \left(1-\cos(2x)\right)\,dx.

Evaluate:

\frac12\left[x - \frac{\sin(2x)}{2}\right]_0^{\pi/2} = \frac12\left(\frac{\pi}{2}\right) = \frac{\pi}{4}.

1.9Test an Improper Integral

Exam I | Problem 1.9 |

Evaluate the improper integral:

\int_1^\infty \frac{1}{x^3}\,dx

Solution

Write it as a limit:

\int_1^\infty \frac{1}{x^3}\,dx = \lim_{b\to\infty}\int_1^b x^{-3}\,dx.

An antiderivative is

\int x^{-3}\,dx = -\frac{1}{2x^2}.

\lim_{b\to\infty}\left[-\frac{1}{2x^2}\right]_1^b = \lim_{b\to\infty}\left(-\frac{1}{2b^2}+\frac12\right) = \frac12.

The integral converges to $\frac12$.

1.10Estimate an Integral with the Trapezoidal Rule

Exam I | Problem 1.10 |

Use the trapezoidal rule with $n=2$ to approximate

\int_0^2 (x^2+1)\,dx.

Solution

With $n=2$,

\Delta x = \frac{2-0}{2} = 1.

The sample points are $x_0=0$, $x_1=1$, and $x_2=2$.

Evaluate the function:

f(0)=1,\qquad f(1)=2,\qquad f(2)=5.

Apply the trapezoidal rule:

\int_0^2 (x^2+1)\,dx \approx \frac{\Delta x}{2}\left[f(0)+2f(1)+f(2)\right].

\frac{1}{2}(1+2\cdot 2+5)=\frac{10}{2}=5.

Difficulty

Integrated

2.1Substitute and Change the Bounds

Exam II | Problem 2.1 |

Evaluate:

\int_0^1 \frac{2x}{1+x^2}\,dx

Solution

Let

u = 1+x^2, \qquad du = 2x\,dx.

Change the bounds:

x=0 \Rightarrow u=1, \qquad x=1 \Rightarrow u=2.

Then

\int_0^1 \frac{2x}{1+x^2}\,dx = \int_1^2 \frac{1}{u}\,du = \left[\ln|u|\right]_1^2 = \ln 2.

2.2Integrate a Logarithm by Parts

Exam II | Problem 2.2 |

Evaluate:

\int_1^e x\ln x\,dx

Solution

Use integration by parts with

u=\ln x,\qquad dv=x\,dx.

Then

du=\frac{1}{x}\,dx,\qquad v=\frac{x^2}{2}.

\int_1^e x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_1^e - \frac12\int_1^e x\,dx.

Now evaluate each part:

\left[\frac{x^2}{2}\ln x\right]_1^e = \frac{e^2}{2}

and

\frac12\int_1^e x\,dx = \frac12\left[\frac{x^2}{2}\right]_1^e = \frac{e^2-1}{4}.

Therefore

\int_1^e x\ln x\,dx = \frac{e^2}{2}-\frac{e^2-1}{4} = \frac{e^2+1}{4}.

2.3Decompose a Rational Function with a Repeated Factor

Exam II | Problem 2.3 |

Evaluate:

\int \frac{1}{x(x+1)^2}\,dx

Solution

Write

\frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}.

Multiply by $x(x+1)^2$:

$$ 1 = A(x+1)^2 + Bx(x+1) + Cx. $$

Expanding gives

$$ 1 = (A+B)x^2 + (2A+B+C)x + A. $$

Match coefficients:

A=1,\qquad A+B=0,\qquad 2A+B+C=0.

A=1,\qquad B=-1,\qquad C=-1.

Therefore

\frac{1}{x(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}.

Integrate term by term:

\int \frac{1}{x(x+1)^2}\,dx = \ln|x| - \ln|x+1| + \frac{1}{x+1} + C.

2.4Use Trig Substitution on a Radical

Exam II | Problem 2.4 |

Compute:

\int_0^{3/2} \frac{dx}{\sqrt{9-x^2}}

Solution

Use the substitution

x = 3\sin\theta, \qquad dx = 3\cos\theta\,d\theta.

The bounds change as follows:

x=0 \Rightarrow \theta=0, \qquad x=\frac32 \Rightarrow \sin\theta=\frac12 \Rightarrow \theta=\frac{\pi}{6}.

Also,

\sqrt{9-x^2} = \sqrt{9-9\sin^2\theta} = 3\cos\theta.

So the integral becomes

\int_0^{\pi/6} \frac{3\cos\theta}{3\cos\theta}\,d\theta = \int_0^{\pi/6} d\theta = \frac{\pi}{6}.

2.5Find the Area Between Two Curves

Exam II | Problem 2.5 |

Find the area between

y = 2x \quad \text{and} \quad y = x^2

on the interval $[0,2]$.

Solution

On $[0,2]$, the line $y=2x$ lies above $y=x^2$.

So the area is

\int_0^2 (2x-x^2)\,dx.

Compute:

\int_0^2 (2x-x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4-\frac{8}{3} = \frac{4}{3}.

2.6Find Volume with the Washer Method

Exam II | Problem 2.6 |

The region between $y=2$ and $y=x$ for $0 \le x \le 2$ is rotated about the $x$-axis. Find the volume.

Solution

Using washers, the outer radius is $R(x)=2$ and the inner radius is $r(x)=x$.

V=\pi\int_0^2 \left(R(x)^2-r(x)^2\right)\,dx =\pi\int_0^2 (4-x^2)\,dx.

Evaluate:

V=\pi\left[4x-\frac{x^3}{3}\right]_0^2 =\pi\left(8-\frac{8}{3}\right) =\frac{16\pi}{3}.

2.7Find Mass from a Density Function

Exam II | Problem 2.7 |

A thin rod has density

\rho(x)=1+2x

for $0 \le x \le 6$.

Find the mass of the rod.

Solution

Mass is the integral of density:

m=\int_0^6 (1+2x)\,dx.

Compute:

m=\left[x+x^2\right]_0^6 = 6+36=42.

2.8Estimate an Integral with Simpson's Rule

Exam II | Problem 2.8 |

Simpson's Rule, Numerical Integration

Use Simpson's rule with $n=2$ to approximate

\int_0^2 x^4\,dx.

Solution

With $n=2$,

\Delta x = \frac{2-0}{2}=1.

The sample points are $0$, $1$, and $2$.

Evaluate the function:

f(0)=0,\qquad f(1)=1,\qquad f(2)=16.

Apply Simpson's rule:

\int_0^2 x^4\,dx \approx \frac{\Delta x}{3}\left[f(0)+4f(1)+f(2)\right].

\frac{1}{3}(0+4\cdot 1+16)=\frac{20}{3}.

Difficulty

Applied

3.1Find Displacement from a Velocity Function

Final | Problem 3.1 |

A particle has velocity

$$ v(t)=3t^2-2t $$

for $0 \le t \le 2$.

Find the displacement over that time interval.

Solution

Displacement is the integral of velocity:

\int_0^2 (3t^2-2t)\,dt.

An antiderivative is

$$ t^3-t^2. $$

Evaluate:

\left[t^3-t^2\right]_0^2 = (8-4)-0 = 4.

So the displacement is $4$.

3.2Find Volume with the Shell Method

Final | Problem 3.2 |

The region under

y=\sqrt{x}

from $x=0$ to $x=4$ is rotated about the $y$-axis. Find the volume.

Solution

Using shells, the radius is $x$ and the height is $\sqrt{x}$.

V = 2\pi \int_0^4 x\sqrt{x}\,dx = 2\pi \int_0^4 x^{3/2}\,dx.

Compute:

\int x^{3/2}\,dx = \frac{2}{5}x^{5/2}.

Therefore

V = 2\pi \left[\frac{2}{5}x^{5/2}\right]_0^4 = \frac{4\pi}{5}\cdot 4^{5/2} = \frac{4\pi}{5}\cdot 32 = \frac{128\pi}{5}.

3.3Find a Probability from a Density Function

Final | Problem 3.3 |

Suppose a random variable has density

$$ f(x)=2x $$

for $0 \le x \le 1$.

Find

P\left(\frac12 \le X \le 1\right).

Solution

Probability is the area under the density:

P\left(\frac12 \le X \le 1\right) = \int_{1/2}^1 2x\,dx.

Compute:

\int_{1/2}^1 2x\,dx = \left[x^2\right]_{1/2}^1 = 1-\frac14 = \frac34.

3.4Evaluate an Improper Integral with a Vertical Asymptote

Final | Problem 3.4 |

Evaluate:

\int_0^1 \frac{1}{\sqrt{x}}\,dx

Solution

This is improper at $x=0$, so write it as a limit:

\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{a\to 0^+}\int_a^1 x^{-1/2}\,dx.

An antiderivative is

2x^{1/2}.

\lim_{a\to 0^+}\left[2\sqrt{x}\right]_a^1 = \lim_{a\to 0^+} (2-2\sqrt{a}) = 2.

3.5Find Geometric Area When the Sign Changes

Final | Problem 3.5 |

Find the geometric area between

$$ f(x)=x^2-4x+3 $$

and the $x$-axis on $[0,4]$.

Solution

First factor:

$$ x^2-4x+3=(x-1)(x-3). $$

So the graph crosses the $x$-axis at $x=1$ and $x=3$.

On $[0,1]$ and $[3,4]$, the function is positive. On $[1,3]$, it is negative.

Thus the geometric area is

\int_0^1 f(x)\,dx - \int_1^3 f(x)\,dx + \int_3^4 f(x)\,dx.

An antiderivative is

\frac{x^3}{3} - 2x^2 + 3x.

Evaluate:

\int_0^1 f(x)\,dx = \frac{4}{3},

\int_1^3 f(x)\,dx = -\frac{4}{3},

\int_3^4 f(x)\,dx = \frac{4}{3}.

So the total area is

\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4.

Difficulty

Challenge

4.1Handle an Endpoint Singularity with Parts

Final | Problem 4.1 |

Evaluate:

\int_0^1 x\ln x\,dx

Solution

This is improper at $x=0$, so treat it as a limit:

\int_0^1 x\ln x\,dx = \lim_{a\to 0^+}\int_a^1 x\ln x\,dx.

Use integration by parts with

u=\ln x,\qquad dv=x\,dx.

Then

du=\frac{1}{x}\,dx,\qquad v=\frac{x^2}{2}.

\int_a^1 x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_a^1 - \frac12\int_a^1 x\,dx.

The boundary term at $x=1$ is $0$, and $a^2\ln a \to 0$ as $a\to 0^+$.

Also,

\frac12\int_a^1 x\,dx = \frac12\left[\frac{x^2}{2}\right]_a^1 = \frac{1-a^2}{4}.

Therefore

\int_0^1 x\ln x\,dx = -\frac14.

4.2Evaluate an Improper Integral After Substitution

Final | Problem 4.2 |

Evaluate:

\int_0^\infty \frac{x}{(1+x^2)^2}\,dx

Solution

Write the integral as a limit:

\int_0^\infty \frac{x}{(1+x^2)^2}\,dx = \lim_{b\to\infty}\int_0^b \frac{x}{(1+x^2)^2}\,dx.

Use the substitution

u = 1+x^2,\qquad du = 2x\,dx.

Then

\int_0^b \frac{x}{(1+x^2)^2}\,dx = \frac12\int_1^{1+b^2} u^{-2}\,du.

Compute:

\frac12\left[-u^{-1}\right]_1^{1+b^2} = \frac12\left(1-\frac{1}{1+b^2}\right).

Now let $b\to\infty$:

\frac12\left(1-\frac{1}{1+b^2}\right)\to \frac12.

So the integral converges to $\frac12$.

4.3Use Shells on a Region Between Curves

Final | Problem 4.3 |

The region enclosed by

y=x \quad \text{and} \quad y=x^2

is rotated about the $y$-axis. Find the volume.

Solution

The curves intersect where

$$ x=x^2, $$

so $x=0$ and $x=1$.

Using shells, the radius is $x$ and the height is

$$ x-x^2. $$

Thus

V = 2\pi\int_0^1 x(x-x^2)\,dx = 2\pi\int_0^1 (x^2-x^3)\,dx.

Evaluate:

V = 2\pi\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 = 2\pi\left(\frac13-\frac14\right) = 2\pi\cdot\frac{1}{12} = \frac{\pi}{6}.

4.4Combine Trig Substitution with a Trig Identity

Final | Problem 4.4 |

Compute:

\int_0^{3/2} \frac{x^2}{\sqrt{9-x^2}}\,dx

Solution

Use the trig substitution

x=3\sin\theta,\qquad dx=3\cos\theta\,d\theta.

The bounds become

x=0 \Rightarrow \theta=0, \qquad x=\frac32 \Rightarrow \theta=\frac{\pi}{6}.

Also,

\sqrt{9-x^2}=3\cos\theta.

So the integral becomes

\int_0^{\pi/6} \frac{9\sin^2\theta}{3\cos\theta}\cdot 3\cos\theta\,d\theta = 9\int_0^{\pi/6}\sin^2\theta\,d\theta.

Use

\sin^2\theta=\frac{1-\cos(2\theta)}{2}.

Then

9\int_0^{\pi/6}\sin^2\theta\,d\theta = \frac{9}{2}\int_0^{\pi/6}\left(1-\cos(2\theta)\right)\,d\theta.

Evaluate:

\frac{9}{2}\left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{\pi/6} = \frac{9}{2}\left(\frac{\pi}{6}-\frac{\sin(\pi/3)}{2}\right).

Since $\sin(\pi/3)=\frac{\sqrt3}{2}$,

\frac{9}{2}\left(\frac{\pi}{6}-\frac{\sqrt3}{4}\right) = \frac{3\pi}{4}-\frac{9\sqrt3}{8}.

Integrals Practice

Selected problems

Quick reference

Direct

1.1Use the Power Rule and Linearity

1.2Differentiate an Accumulation Function

1.3Evaluate a Definite Integral with the FTC

1.4Find an Average Value

1.5Use Substitution on a Composite Function

1.6Use Integration by Parts on a Product

1.7Decompose a Simple Rational Integral

1.8Use a Trig Identity Before Integrating

1.9Test an Improper Integral

1.10Estimate an Integral with the Trapezoidal Rule

Integrated

2.1Substitute and Change the Bounds

2.2Integrate a Logarithm by Parts

2.3Decompose a Rational Function with a Repeated Factor

2.4Use Trig Substitution on a Radical

2.5Find the Area Between Two Curves

2.6Find Volume with the Washer Method

2.7Find Mass from a Density Function

2.8Estimate an Integral with Simpson's Rule

Applied

3.1Find Displacement from a Velocity Function

3.2Find Volume with the Shell Method

3.3Find a Probability from a Density Function

3.4Evaluate an Improper Integral with a Vertical Asymptote

3.5Find Geometric Area When the Sign Changes

Challenge

4.1Handle an Endpoint Singularity with Parts

4.2Evaluate an Improper Integral After Substitution

4.3Use Shells on a Region Between Curves

4.4Combine Trig Substitution with a Trig Identity