# AP Calculus AB — Indefinite & Definite Integrals

This is the Calculus AB unit quiz for indefinite and definite integrals.

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 Question 1
Evaluate $∫(4x^7 + 5x^3 + 7x + 5)dx$.

 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 $∫x^ndx = \frac{x^{n + 1}}{n + 1} + C$. 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 $∫\dfrac{ln⁡(x)}{x}dx$.

 A $\dfrac{1}{2}(ln(x))^2 + C$ B $\dfrac{1}{x^2} + C$ C $2ln(x) + C$ 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

$u = ln(x)$. $\frac{du}{dx} = \frac{1}{x}$. $du = \frac{1}{x}dx$.

Substituting $u$ into the integral gives $∫ u du = \frac{1}{2}(u)^2 + C$.

Since $u = ln(x)$, the integral is equal to $\frac{1}{2}(ln(x))^2 + C$.
 Question 3
Evaluate $∫6x^2cos⁡(x^3)dx$.

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

$\frac{du}{dx} = 3x^2$. $du = 3x^2$.

The integral can then be simplified to $2∫(cos⁡(u))du$ $= 2sin(u) + C$.

Replacing $u$ with $x^3$ gives $2sin⁡(x^3 ) + C$.
 Question 4
Evaluate $∫(tan⁡(x) + sec^2⁡(x))dx$.

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

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

 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

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

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:

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

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 $∫e^{x + 4}dx$.

 A $\dfrac{1}{4}e^{x + 4} + C$ B $\dfrac{1}{2}e^{x + 4} + C$ 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 $∫(e^u)du = e^u + C$.

Since $u = x + 4$, the correct answer is $e^{x + 4} + C$.
 Question 7
Evaluate $∫\left(dfrac{−e^{−x}}{1 + e^{−x}}\right)dx$.

 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 $ln(|1 + e^{−x}|) + C$.
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