Definite and Indefinite Integrals Questions and Answers

Answer: B

The value is \(p = \frac{-1}{3}\).

Substitute \(u = \frac{1}{x^{6}} + 1\), so \(du = -\frac{6}{x^{7}}dx\) and \(\frac{dx}{x^{7}} = -\frac{du}{6}\).

\[ \begin{aligned} \int \frac{dx}{x^{7}u^{2/3}} &= -\frac{1}{6}\int u^{-2/3}du \\ &= -\frac{1}{6}\cdot 3u^{1/3} = -\frac{1}{2}u^{1/3} \end{aligned} \]

The answer is written in terms of \(\frac{1}{u}\), and \(u^{1/3} = \left(\frac{1}{u}\right)^{-1/3}\), so \(p = -\frac{1}{3}\). The value \(\frac{1}{3}\) is the power of u itself, before rewriting in terms of the reciprocal.

Answer: C

The integral is \(\frac{-1}{9}(secx + tanx)^{-9} + C\).

Let \(t = \sec x + \tan x\). Then \(\frac{dt}{dx} = \sec x\tan x + \sec^{2}x = \sec x\,(\sec x + \tan x) = t\sec x\), so \(\sec x\,dx = \frac{dt}{t}\).

\[ \begin{aligned} \int \frac{\sec x\,dx}{t^{9}} &= \int \frac{dt}{t^{10}} \\ &= -\frac{1}{9}t^{-9} + C \end{aligned} \]

The positive \(\frac{1}{9}\) misses the minus sign from integrating a negative power, and the options with power +9 integrate as if t were in the numerator.

Answer: E

The integral is \(\log|9e^{x} - 4e^{-x}| + C\).

The numerator is exactly the derivative of the denominator: \(\frac{d}{dx}(9e^{x} - 4e^{-x}) = 9e^{x} + 4e^{-x}\). So the integrand has the form \(\frac{f'(x)}{f(x)}\).

\[ \int \frac{f'(x)}{f(x)}dx = \log|f(x)| + C \]

Using the numerator inside the logarithm, \(\log|9e^{x} + 4e^{-x}|\), would differentiate to the reciprocal of the integrand's ratio. The options without a logarithm, and the one with 4 and 9 swapped, do not differentiate back to the integrand.

Answer: A

The integral is \(e^{2\theta}cos^{2}\theta + C\).

Look for a product whose derivative gives the integrand. With \(f(\theta) = \cos^{2}\theta\), \(f'(\theta) = -2\cos\theta\sin\theta = -\sin 2\theta\).

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

This is the pattern \(\int e^{kx}(kf + f')dx = e^{kx}f + C\) with k = 2. \(2e^{2\theta}\cos^{2}\theta\) has twice the derivative, and the options with \(\sin 2\theta\), \(\sin\theta\) or \(\cos 2\theta\) differentiate to other expressions.

Answer: B

The integral is \(e^{\left(x + \frac{1}{x}\right)} + C\).

The factor \(\frac{x^{2} - 1}{x^{2}} = 1 - \frac{1}{x^{2}}\) is the derivative of the exponent \(x + \frac{1}{x}\). Put \(t = x + \frac{1}{x}\):

\[ \int e^{t}\,dt = e^{t} + C = e^{x + \frac{1}{x}} + C \]

The options multiplying by x or \(x^{2}\), or adding x or \(x^{2}\), would produce extra terms on differentiation that are not in the integrand.

Answer: C

The area is \(\frac{1}{2}\) square units.

On \(1 \le x \le 2\) the line y = x - 1 is above the x-axis, rising from 0 to 1.

\[ \begin{aligned} \int_{1}^{2}(x - 1)\,dx &= \left[\frac{(x - 1)^{2}}{2}\right]_{1}^{2} \\ &= \frac{1}{2} \end{aligned} \]

Geometrically it is a right triangle with base 1 and height 1, area \(\frac{1}{2}\). 1 is the area of the enclosing square, and 2 or 4 integrate y = x or use the wrong limits.

Answer: E

The value is \(\frac{\pi}{4} - \frac{1}{2}log2\).

Use the property \(\int_{0}^{a}f(t)\,dt = \int_{0}^{a}f(a - t)\,dt\) with a = 1. Replacing t by 1 - t maps the interval [0, 1] onto itself.

\[ \begin{aligned} \int_{0}^{1}\tan^{-1}(1 - t)\,dt &= \int_{0}^{1}\tan^{-1}(t)\,dt \\ &= \frac{\pi}{4} - \frac{1}{2}\log 2 \end{aligned} \]

The integral is unchanged, so the answer is the given value. The option with + log 2 changes the sign, the one with log 3 changes the logarithm, and \(\frac{\pi}{2}\) doubles the first term.

Answer: D

The value is 1.

Multiply numerator and denominator by \(1 - \sin x\).

\[ \begin{aligned} \frac{1}{1 + \sin x} &= \frac{1 - \sin x}{\cos^{2}x} = \sec^{2}x - \sec x\tan x \\ \int \frac{dx}{1 + \sin x} &= \tan x - \sec x = \frac{\sin x - 1}{\cos x} \end{aligned} \]

As \(x \to \frac{\pi}{2}\), \(\frac{\sin x - 1}{\cos x} \to 0\); at x = 0 it equals -1. So the integral is 0 - (-1) = 1. The integrand lies between \(\frac{1}{2}\) and 1 on an interval of length \(\frac{\pi}{2}\), so values like 2, \(\frac{1}{4}\) or 0 are impossible.

Answer: E

The value is 12.

On [-2, 2], x + 3 is always positive (between 1 and 5), so \(|x + 3| = x + 3\).

\[ \begin{aligned} \int_{-2}^{2}(x + 3)\,dx &= \left[\frac{x^{2}}{2} + 3x\right]_{-2}^{2} \\ &= 8 - (-4) = 12 \end{aligned} \]

The x term integrates to zero over the symmetric interval, leaving \(3 \times 4 = 12\). 8 keeps only the upper limit and 16 adds instead of subtracting at the lower limit.

Answer: E

The integral is \(2\sin x + C\).

Simplify first: \(\frac{\sin 2x}{\sin x} = \frac{2\sin x\cos x}{\sin x} = 2\cos x\), and \(\int 2\cos x\,dx = 2\sin x + C\).

\(\sin x + C\) loses the factor 2, and \(2\cos x\) or \(-\cos x\) are what you get by differentiating instead of integrating.