Definite and Indefinite Integrals Questions and Answers

Answer: A

The integral is 1.

Write \(1 + \sin 2x = \sin^{2}x + \cos^{2}x + 2\sin x\cos x = (\sin x + \cos x)^{2}\). On \([0, \frac{\pi}{4}]\), \(\sin x + \cos x \gt 0\), so the square root is \(\sin x + \cos x\).

\[ \begin{aligned} \int_{0}^{\pi/4}(\sin x + \cos x)dx &= [-\cos x + \sin x]_{0}^{\pi/4} \\ &= \left(-\tfrac{1}{\sqrt{2}} + \tfrac{1}{\sqrt{2}}\right) - (-1 + 0) = 1 \end{aligned} \]

\(\sqrt{2} - 1\) and \(\sqrt{2} + 1\) come from sign slips in the antiderivative of \(\sin x\), and the negative values are impossible for an integral of a positive function.

Answer: A

The integral is \(\log|\sin x| + C\).

Simplify the integrand first. \(\tan x + \cot x = \frac{1}{\sin x\cos x}\) and \(1 + \tan^{2}x = \sec^{2}x\).

\[ \begin{aligned} \frac{\tan x + \cot x}{\sec^{2}x} &= \frac{\cos^{2}x}{\sin x\cos x} = \cot x \\ \int \cot x\,dx &= \log|\sin x| + C \end{aligned} \]

\(\log|\cos x|\) would come from integrating \(-\tan x\), and \(\log|\tan x|\) from integrating \(\frac{1}{\sin x\cos x}\) without the \(\sec^{2}x\) in the denominator.

Answer: D

The integral is \((x\log x)e^{-x} + C\).

Check by differentiating the keyed answer with the product rule.

\[ \begin{aligned} \frac{d}{dx}\left[x\log x\,e^{-x}\right] &= (\log x + 1)e^{-x} - x\log x\,e^{-x} \\ &= e^{-x}(1 + (1 - x)\log x) \end{aligned} \]

This is the integrand, so the answer is confirmed. \(e^{x}\) in place of \(e^{-x}\) would produce \(1 + (1 + x)\log x\), and the \(x^{2}\) or \(e^{-2x}\) versions give different derivatives.

Answer: D

The integral is \(\frac{\sin x}{x} + C\).

The integrand is the quotient-rule derivative of \(\frac{\sin x}{x}\).

\[ \begin{aligned} \frac{d}{dx}\left(\frac{\sin x}{x}\right) \\ = \frac{x\cos x - \sin x}{x^{2}} \end{aligned} \]

Differentiating \(\frac{\cos x}{x}\) would give \(\frac{-x\sin x - \cos x}{x^{2}}\), and \(\frac{\sin x}{x^{2}}\) would give a cubic in the denominator, so neither matches.

Answer: B

The integral is \(x^{2}\log x + C\).

Differentiate the keyed answer with the product rule.

\[ \begin{aligned} \frac{d}{dx}(x^{2}\log x) &= 2x\log x + x^{2} \cdot \frac{1}{x} \\ &= x(1 + 2\log x) \end{aligned} \]

This is exactly the integrand. \(x\log x\) differentiates to \(\log x + 1\), and \(x^{3}\log x\) to \(3x^{2}\log x + x^{2}\), so neither works. Integration by parts gives the same: \(\int 2x\log x\,dx = x^{2}\log x - \frac{x^{2}}{2}\), and adding \(\int x\,dx = \frac{x^{2}}{2}\) leaves \(x^{2}\log x\).

Answer: A

The value is -1.

\(\frac{|x|}{x}\) is -1 for x < 0 and +1 for x > 0, so split at 0.

\[ \begin{aligned} \int_{-2}^{1}\frac{|x|}{x}dx &= \int_{-2}^{0}(-1)\,dx + \int_{0}^{1}dx \\ &= -2 + 1 = -1 \end{aligned} \]

Treating \(\frac{|x|}{x}\) as 1 throughout gives 3, the length of the interval, and treating it as -1 throughout gives -3. Geometrically, the integrand is a step: the area below the axis on [-2, 0] is 2 and the area above it on [0, 1] is 1.

Answer: D

The value is \(\pi - 2\).

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

\[ \begin{aligned} \int_{0}^{\pi}\frac{dx}{1 + \sin x} &= \int_{0}^{\pi}\frac{1 - \sin x}{\cos^{2}x}dx \\ &= \left[\tan x - \sec x\right]_{0}^{\pi} = 2 \\ \text{value} &= \pi - 2 \end{aligned} \]

The antiderivative \(\tan x - \sec x\) equals 1 at \(\pi\) and -1 at 0 (it is continuous across \(\frac{\pi}{2}\), with limit 0 there). \(\pi + 2\) adds instead of subtracting the second part.

Answer: B

The value is 15.

The greatest integer \([x]\) is 0, 1 and 2 on [0, 1), [1, 2) and [2, 3).

\[ \begin{aligned} \int_{0}^{3}x^{2}[x]\,dx &= 0 + \int_{1}^{2}x^{2}dx + 2\int_{2}^{3}x^{2}dx \\ &= \tfrac{7}{3} + 2 \times \tfrac{19}{3} = \tfrac{45}{3} = 15 \end{aligned} \]

Without the factor [x] the integral would be 9. Forgetting the 2 on the last piece gives \(\frac{26}{3}\), and treating [x] as x gives \(\frac{81}{4}\).

Answer: D

The value is \(\frac{\pi}{4}\).

Call the integral I. Replacing x by \(\frac{\pi}{2} - x\) swaps sine and cosine.

\[ \begin{aligned} I &= \int_{0}^{\pi/2}\frac{\sin^{11}x}{\sin^{11}x + \cos^{11}x}dx \\ 2I &= \int_{0}^{\pi/2}1\,dx = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \end{aligned} \]

Adding the two forms of I gives the integral of 1. \(\frac{\pi}{2}\) is 2I, forgetting to halve. The power 11 plays no part.