Definite and Indefinite Integrals Questions and Answers

Answer: A

The integral is \(\frac{x^{3}}{3}\ln x - \frac{x^{3}}{9} + C\).

Integrate by parts with \(u = \ln x\) and \(dv = x^{2}\,dx\), so \(du = \frac{dx}{x}\) and \(v = \frac{x^{3}}{3}\), writing \(I\) for the integral:

\[ \begin{aligned} I &= \frac{x^{3}}{3}\ln x - \int \frac{x^{2}}{3}\,dx \\ &= \frac{x^{3}}{3}\ln x - \frac{x^{3}}{9} + C \end{aligned} \]

Check by differentiating: \(x^{2}\ln x + \frac{x^{2}}{3} - \frac{x^{2}}{3} = x^{2}\ln x\). The option with \(+\frac{x^{3}}{9}\) drops the minus sign of integration by parts. The options with \(\frac{x^{3}}{9}\ln x\) put the wrong coefficient on the log term.

Answer: B

The integral equals \(\frac{1}{3}\).

Split the integrand so that one part has its numerator equal to the derivative of its denominator (up to a constant):

\[ \begin{aligned} \frac{1}{3^{x} + 3} &= \frac{1}{3}\cdot\frac{3}{3^{x} + 3} \\ &= \frac{1}{3}\left(1 - \frac{3^{x}}{3^{x} + 3}\right) \end{aligned} \]

Since \(\frac{d}{dx}(3^{x} + 3) = 3^{x}\ln 3\), the second part integrates to a logarithm:

\[ \begin{aligned} \int_{0}^{2} \frac{3^{x}}{3^{x} + 3}\,dx \\ &= \frac{1}{\ln 3}\Big[\ln(3^{x} + 3)\Big]_{0}^{2} \\ &= \frac{\ln 12 - \ln 4}{\ln 3} = 1 \end{aligned} \]

So the integral is \(\frac{1}{3}(2 - 1) = \frac{1}{3}\).

The logarithm cancels exactly because \(\ln 12 - \ln 4 = \ln 3\), so the options containing \(\log_{e} 3\) keep a factor that disappears. \(\frac{1}{2}\) is too large: the integrand never exceeds \(\frac{1}{4}\) on \([0, 2]\), so the integral is below \(\frac{1}{2}\).

Answer: D

The integral is \(2x + log_{e}|x^{2} + x + 1| + C\).

Divide first: \(2x^{2} + 4x + 3 = 2(x^{2} + x + 1) + (2x + 1)\), so

\[ \begin{aligned} &\frac{2x^{2} + 4x + 3}{x^{2} + x + 1} \\ &= 2 + \frac{2x + 1}{x^{2} + x + 1} \end{aligned} \]

The second fraction has its numerator equal to the derivative of its denominator, so it integrates to \(\log_{e}|x^{2} + x + 1|\), and the 2 integrates to \(2x\). Check by differentiating: \(2 + \frac{2x + 1}{x^{2} + x + 1}\). Option E swaps the coefficients (\(x + 2\log\)), and options A to C omit the polynomial part.

Answer: A

The integral is \(\frac{1}{2}(sin^{-1}x)^{2} + C\).

Put \(t = \sin^{-1}x\), so \(dt = \frac{dx}{\sqrt{1 - x^{2}}}\):

\[ \begin{aligned} \int\frac{\sin^{-1}x}{\sqrt{1 - x^{2}}}\,dx &= \int t\,dt \\ &= \frac{t^{2}}{2} + C \end{aligned} \]

Option E forgets the \(\frac{1}{2}\) (its derivative is twice the integrand). The options with \(\sqrt{1 - x^{2}}\) look like integration by parts but differentiate to something else entirely.

Answer: E

The integral is \(\frac{1}{2}(x^{8} + 1)^{1/4} + C\).

Put \(u = x^{8} + 1\), so \(du = 8x^{7}\,dx\):

\[ \begin{aligned} \int x^{7}(x^{8} + 1)^{-3/4}dx &= \frac{1}{8}\int u^{-3/4}du \\ &= \frac{1}{8}\cdot 4u^{1/4} + C = \frac{1}{2}u^{1/4} + C \end{aligned} \]

\(4(x^{8} + 1)^{1/4}\) forgets the factor \(\frac{1}{8}\), and \((x^{8} + 1)^{1/4}\) applies it only partly. The forms with \(1 + \frac{1}{x^{8}}\) belong to a different integrand, \(x^{-9}(1 + x^{-8})^{-3/4}\).

Answer: C

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

Expand the integrand as \(e^{x}(\sec x + \sec x\tan x)\). This has the form \(e^{x}\left(f(x) + f'(x)\right)\) with \(f(x) = \sec x\), since \(\frac{d}{dx}\sec x = \sec x\tan x\), and

\[ \int e^{x}\left(f(x) + f'(x)\right)dx = e^{x}f(x) + C \]

Check: \(\frac{d}{dx}(e^{x}\sec x) = e^{x}\sec x + e^{x}\sec x\tan x\). \(e^{x}\tan x\) would need \(\tan x + \sec^{2}x\) in the bracket, and \(e^{x}\sec x\tan x\) differentiates to something with \(\sec^{3}x\).

Answer: A

The integral is \(sin\left(e^{x}(x^{2} - 2x)\right) + C\).

Let \(t = e^{x}(x^{2} - 2x)\). By the product rule:

\[ \begin{aligned} \frac{dt}{dx} &= e^{x}(x^{2} - 2x) + e^{x}(2x - 2) \\ &= e^{x}(x^{2} - 2) \end{aligned} \]

This is exactly the factor outside the cosine, so the integral is \(\int\cos t\,dt = \sin t + C\). The options with \(x^{2} - 2\) inside the sine put the wrong expression inside, and those with an extra \(e^{x}\) or \(x^{2}e^{x}\) in front would differentiate to extra terms.

Answer: D

\(\alpha + \beta = \) 28.

An antiderivative is \(F(x) = -2x^{3} + 18x\).

\[ \begin{aligned} F(1) &= -2 + 18 = 16 \\ F(-\sqrt{3}) &= -2(-3\sqrt{3}) - 18\sqrt{3} = -12\sqrt{3} \\ F(1) - F(-\sqrt{3}) &= 16 + 12\sqrt{3} \end{aligned} \]

So \(\alpha = 16\), \(\beta = 12\) and \(\alpha + \beta = 28\). A sign slip in \((-\sqrt{3})^{3} = -3\sqrt{3}\) gives \(16 + 24\sqrt{3}\) or \(16 - 12\sqrt{3}\), which lead to 40 or 4, neither offered.

Answer: C

The integral is \(\frac{3\pi}{20}\).

The limits add to \(\frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{2}\), so use \(\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx\). Replacing \(x\) by \(\frac{\pi}{2} - x\) turns \(\cot x\) into \(\tan x\):

\[ \begin{aligned} I &= \int \frac{\cot^{3}x}{1 + \cot^{3}x}dx = \int \frac{\tan^{3}x}{1 + \tan^{3}x}dx \\ 2I &= \int_{\pi/10}^{2\pi/5} 1\,dx = \frac{3\pi}{10} \end{aligned} \]

(The two integrands add to 1, since \(\frac{\tan^{3}x}{1 + \tan^{3}x} = \frac{1}{\cot^{3}x + 1}\).) So \(I = \frac{3\pi}{20}\). \(\frac{3\pi}{10}\) is the length of the interval without halving; \(\frac{\pi}{4}\) is the familiar answer for limits 0 to \(\frac{\pi}{2}\).

Answer: C

The area is \(\frac{3}{14}\) square units.

\(x^{5/2}\) is defined for \(x \geq 0\), and the curves meet where \(x^{5/2} = x\), at \(x = 0\) and \(x = 1\). Between them the line lies above the curve:

\[ \begin{aligned} A &= \int_{0}^{1}\left(x - x^{5/2}\right)dx \\ &= \frac{1}{2} - \frac{2}{7} = \frac{3}{14} \end{aligned} \]

\(\frac{2}{7}\) is the area under the curve alone, and \(\frac{3}{7}\) doubles the answer as if the region were reflected into \(x \lt 0\), where the curve does not exist.