AP Calculus AB — Indefinite & Definite Integrals

Below is our AP Calculus AB unit test on indefinite and definite integrals. These questions cover properties of integrals, basic anti-derivatives, u-substitution, trig integrals, and definite integrals. Integration is a large part of the AP exam and understanding how the anti-derivative works will become a very important mathematical tool in the future.

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 Question 1
Evaluate:

$\displaystyle\int (4x^7 + 5x^3 + 7x + 5)dx$

 A $\frac{1}{2}x^8 + \frac{5}{4}x^4 + \frac{7}{2}x^2 +5x$ B $\frac{1}{2}x^8 + \frac{5}{4}x^4 + \frac{7}{2}x^2 +5x + C$ C $\frac{1}{2}x^8 + \frac{5}{4}x^4 + \frac{7}{2}x^2 + C$ D $\frac{1}{2}x^8 + C$
Question 1 Explanation:
The correct answer is (B). This question requires the usage of the formula:

$\displaystyle\int x^ndx = \frac{x^{n + 1}}{n + 1} + C$

Applying this formula to each term of the integrand gives:

$\frac{1}{2}x^8 + \frac{5}{4}x^4 + \frac{7}{2}x^2 + 5x + C$
 Question 2
Evaluate:

$\displaystyle\int \frac{\ln{⁡(x)}}{x}dx$

 A $\dfrac{1}{2}(\ln(x))^2 + C$ B $\dfrac{1}{x^2} + C$ C $2\ln(x) + C$ D $\ln(x) + \dfrac{1}{x^2} + C$
Question 2 Explanation:
The correct answer is (A). This integral requires a u-substitution. We know that the derivative of $\ln(x)$ is $\frac{1}{x}$, so let $u = \ln(x)$:

$\dfrac{du}{dx} = \dfrac{1}{x}$

$du = \dfrac{1}{x}dx$

Substituting $u$ into the integral gives:

$\displaystyle\int u~du = \frac{1}{2}(u)^2 + C$

Since $u = \ln(x)$, the integral is equal to:

$\dfrac{1}{2}(\ln(x))^2 + C$
 Question 3
Evaluate:

$\displaystyle\int6x^2\cos⁡(x^3)dx$

 A $2 \cos(x^3) + C$ B $6\sin(x^3) + C$ C $2\sin(x^3) + C$ D $6\cos(x^3) + C$
Question 3 Explanation:
The correct answer is (C). This integral requires a u-substitution. Let $u = x^3$, since $x^3$ is the inner function:

$\dfrac{du}{dx} = 3x^2$

$du = 3x^2$

The integral can then be simplified to:

$2\displaystyle\int(\cos⁡(u))du$ $= 2\sin(u) + C$

Replacing $u$ with $x^3$ gives: $2\sin⁡(x^3 ) + C$
 Question 4
Evaluate:

$\displaystyle\int(\tan⁡(x) + \sec^2⁡(x))dx$

 A $–\ln |\sin(x)| + \tan(x) + C$ B $–\ln |\cos(x)| + \tan(x) + C$ C $\frac{1}{2}\tan^2(x) + \frac{1}{3}\sec^3(x) + C$ D $\cos(x) + \frac{1}{3}\sec^3(x) + C$
Question 4 Explanation:
The correct answer is (B). This integral can be split into:

$\displaystyle\int(\tan⁡(x))dx + \displaystyle\int(\sec^2⁡(x))dx$

The solution to the first integral is: $–\ln {|\cos(x)|} + C$

The solution to the second integral is: $\tan(x) + C$

The sum of these two results is: $–\ln {|\cos(x)|} + \tan(x) + C$
 Question 5
Evaluate:

$\displaystyle\int\left(\dfrac{5x}{9 + (x^2)^2}\right)dx$

 A $\dfrac{5}{6}\arctan\left(\dfrac{x^2}{3}\right) + C$ B $\dfrac{5}{3}\arctan\left(\dfrac{x^2}{3}\right) + C$ C $\dfrac{5}{6}\arctan\left(\dfrac{x^2}{9}\right) + C$ D $\dfrac{5}{6}\arctan\left(\dfrac{x}{3}\right) + C$
Question 5 Explanation:
The correct answer is (A). This integral requires the memorization of the formula

$\displaystyle\int\left(\dfrac{du}{u^2 + a^2}\right)$ $= \dfrac{1}{a}\arctan\left(\dfrac{u}{a}\right) + C$

For this integral, we need a $2x$ on the numerator, since the derivative of $x^2$ is $2x$. This can be done by factoring out $\frac{5}{2}$ from the numerator as follows:

$\dfrac{5}{2}\displaystyle\int\left(\dfrac{2x}{(x^2)^2 + (3)^2}\right)dx$

We see that $u = x^2$ and $a = 3$. Plugging these values into the given formula results in the following:

$\dfrac{5}{2} \cdot \dfrac{1}{3}\arctan\left(\dfrac{x^2}{3}\right) + C$ $= \dfrac{5}{6}\arctan\left(\dfrac{x^2}{3}\right) + C$
 Question 6
Evaluate:

$\displaystyle\int e^{x + 4}dx$

 A $\frac{1}{4}e^{x + 4} + C$ B $\frac{1}{2}e^{x + 4} + C$ C $e^{x + 4} + C$ D $\ln(x)e^{x + 4} + C$
Question 6 Explanation:
The correct answer is (C). This integral can be solved using $u$-substitution. Let $u = x + 4$, $du = dx$.

This gives: $\displaystyle\int(e^u)du = e^u + C$

Since $u = x + 4$, the correct answer is $e^{x + 4} + C$.
 Question 7
Evaluate:

$\displaystyle\int \left(\dfrac{−e^{−x}}{1 + e^{−x}}\right)dx$

 A $−e^{−x} + C$ B $x − e^{−x} + C$ C $x \ln(|x|) − e^{−x} + C$ D $\ln(|1 + e^{−x}|) + C$
Question 7 Explanation:
The correct answer is (D). Examining the integral, we can see that it is in the form $\frac{u'}{u}$ since the derivative of $u = 1 + e^{−x}$ $= −e^{−x}$.

Since the integral of any function in the form $\frac{u'}{u} = \ln(|u|) + C$, the correct answer is $\ln(|1 + e^{−x}|) + C$.
 Question 8
Evaluate:

$\displaystyle\int_0^5 (3x^2 + 7x + 4) dx$

 A $0$ B $54$ C $169$ D $232.5$
Question 8 Explanation:
The correct answer is (D). First, we need to solve the indefinite integral, then we can plug in the given bounds. Using the power rule, the integral is equal to $x^3 + \frac{7}{2}x^2 + 4x$ on the interval $[0, 5]$. Evaluating the definite integral using the first fundamental theorem gives:

$\left[5^3 + \dfrac{7}{2}(5^2) + 4(5) \right]$ $− \left[0^3 + \dfrac{7}{2}(0)^2 + 4(0) \right] = 232.5$
 Question 9
Evaluate:

$\displaystyle\int_0^7 3x^2 e^{x^3} dx$

 A $e^{147} − 1$ B $e^{243} − 1$ C $e^{512} − 1$ D $e^{1701} − 1$
Question 9 Explanation:
The correct answer is (B). First, we can solve the integral through u-substitution. Let $u = x^3$, $du = 3x^2 dx$. The integral of $e^u = e^u$.

Since $u = x^3$, the definite integral is equal to $e^{x^3}$ on the interval $[0, 7]$.

$e^{7^3} − e^{0^3} = e^{243} − 1$
 Question 10
Evaluate:

$\displaystyle\int_0^{2π} \tan(x) dx$

 A $0$ B $\ln |1|$ C $−\ln |1|$ D $\tan(2π) − \tan(0)$
Question 10 Explanation:
The correct answer is (A). Start by evaluating the indefinite integral. The anti-derivative of $\tan⁡(x)$ is $−\ln ⁡|\cos⁡(x)|$. Since we are evaluating a definite integral on the interval $[0, 2π]$, the constant of integration can be ignored.

$−\ln ⁡|\cos⁡(0)| − (−\ln ⁡|\cos⁡(2π)| )$

$= −\ln ⁡|1| + \ln⁡ |1| = 0$.
 Question 11
Evaluate:

$\displaystyle\int_{−1}^1 \dfrac{1}{1 + x^2} dx$

 A $\dfrac{π}{4}$ B $\dfrac{π}{2}$ C $\dfrac{π}{3}$ D $\dfrac{π}{6}$
Question 11 Explanation:
The correct answer is (B). Using the anti-derivative formula for $\arctan(x)$, the indefinite integral is equal to $\arctan⁡(x)$ on the interval $[−1, 1]$. Evaluating the definite integral gives:

$\arctan⁡(1) − \arctan⁡(−1)$ $= \dfrac{π}{4} − \left(− \dfrac{π}{4}\right) = \dfrac{π}{2}$
 Question 12
Evaluate:

$\displaystyle\int_{−5}^2 (e^x + 5) dx$

 A $e^2 − e^{−5} − 35$ B $e^2 − e^{−5} + 35$ C $e^{−2} − e^{5} − 35$ D $e^{−2} + e^{5} − 35$
Question 12 Explanation:
The correct answer is (B). Begin by evaluating the indefinite integral. This results in $e^x + 5x$. Evaluating on the interval $[−5, 2]$ gives:

$[e^2 + 5(2)] − [e^{−5} + 5(−5)]$ $= e^2 − e^{−5} + 35$
 Question 13
Evaluate:

$\displaystyle\int_1^{e^2} \left(\dfrac{1}{x} + 2 \right) dx$

 A $2 + 2e^2$ B $2 − 2e^2$ C $4 + 2e^2$ D $2e^2$
Question 13 Explanation:
The correct answer is (D). The evaluation of the indefinite integral is $\ln⁡|x| + 2x$. Since the bounds of the definite integral are $[1, e^2 ]$, applying the first fundamental theorem of calculus gives:

$[\ln ⁡|e^2| + 2(e^2 )] − [\ln⁡|1| + 2(1)]$ $= 2 + 2e^2 − 2 = 2e^2$
 Question 14
Evaluate:

$\displaystyle\int_0^{\frac{π}{2}} \sin(2x + π)$

 A $−1$ B $0$ C $1$ D $\dfrac{π}{2}$
Question 14 Explanation:
The correct answer is (A). Evaluating the indefinite integral can be done using a u-substitution. Let $u = 2x + π$. Then $\frac{du}{dx} = 2$ and $\frac{1}{2} du = dx$.

Substituting into the integral gives:

$\frac{1}{2}\displaystyle\int (\sin⁡(u) du = −\frac{1}{2} \cos⁡(u)$ $= −\frac{1}{2} \cos⁡(2x + π)$.

Evaluating the definite integral on the bounds $[0, \frac{π}{2}]$ using the first fundamental theorem gives:

$\left[−\dfrac{1}{2} \cos⁡ \left(2 \left(\dfrac{π}{2}\right) + π \right) \right]~−$ $\left[−\dfrac{1}{2} \cos⁡(2(0)+π) \right]$

$= −\dfrac{1}{2} − \left(\dfrac{1}{2} \right) = −1$
 Question 15
Evaluate:

$\displaystyle\int_2^5 (x^2 e^2) dx$

 A $\dfrac{100}{3}e^2$ B $\dfrac{117}{3}e^2$ C $\dfrac{125}{3}e^2$ D $\dfrac{133}{3}e^2$
Question 15 Explanation:
The correct answer is (B). Evaluating the indefinite integral using the power rule gives $\frac{1}{3} x^3 e^2$, ignoring the constant of integration. Evaluating the definite integral on the interval $[2, 5]$ gives:

$\dfrac{1}{3} e^2 [5^3 − 2^3 ] = \dfrac{1}{3} e^2 (117)$ $= \dfrac{117}{3} e^2$
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