Definite and Indefinite Integrals Questions and Answers

Answer: A

The integral is \(\frac{log_{e}5}{2log_{e}3}\).

Put \(u = 3^{2x} + 1\), so \(du = 2\ln 3 \cdot 3^{2x}dx\):

\[ \begin{aligned} \int_{0}^{1}\frac{3^{2x}}{3^{2x} + 1}dx &= \frac{1}{2\ln 3}\left[\ln(3^{2x} + 1)\right]_{0}^{1} \\ &= \frac{\ln 10 - \ln 2}{2\ln 3} = \frac{\ln 5}{2\ln 3} \end{aligned} \]

The other denominators (3 or 9 times \(\ln 3\)) come from differentiating \(3^{2x}\) wrongly, for example as \(3^{2x}\ln 9\) and then again multiplying by 2 or 3.

Answer: E

The integral is \(\frac{1}{4}\log_{e}\left|\frac{x^{4}}{1 + x^{4}}\right| + C\).

Multiply top and bottom by \(x^{3}\) and put \(t = x^{4}\), \(dt = 4x^{3}dx\):

\[ \begin{aligned} \int\frac{x^{3}}{x^{4}(1 + x^{4})}dx &= \frac{1}{4}\int\frac{dt}{t(1 + t)} \\ &= \frac{1}{4}\int\left(\frac{1}{t} - \frac{1}{1 + t}\right)dt \\ &= \frac{1}{4}\log\left|\frac{t}{1 + t}\right| + C \end{aligned} \]

The options with \(\frac{x}{1 + x^{4}}\) put \(x\) where \(x^{4}\) belongs, and the ones with \(\log|1 + x^{4}|\) alone lose the \(\frac{1}{t}\) part.

Answer: E

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

Since \(\sin 4\theta = 2\sin 2\theta\cos 2\theta\), the integrand is \(2\cos 2\theta\):

\[ \int 2\cos 2\theta\,d\theta = \sin 2\theta + C \]

\(2\sin 2\theta\) forgets to divide by the 2 from the chain rule, \(\cos 2\theta\) is a derivative of the wrong kind, and the options in \(\sin\theta\) or \(\cos\theta\) do not differentiate back to \(2\cos 2\theta\).

Answer: D

The integral is \(\tan(\sqrt{2x + 5}) + C\).

Put \(u = \sqrt{2x + 5}\); then \(du = \frac{dx}{\sqrt{2x + 5}}\), exactly the rest of the integrand:

\[ \int\sec^{2}u\,du = \tan u + C = \tan\sqrt{2x + 5} + C \]

The factors 2 and \(\frac{1}{2}\) cancel inside \(du\), so the options with 2 or \(\frac{1}{2}\) in front are wrong, and those in \(\tan(2x + 5)\) lose the square root.

Answer: D

The integral is \(-\frac{1}{2}\log_{e}|5 - x^{2}| + C\).

The derivative of \(5 - x^{2}\) is \(-2x\), so write the numerator as \(-\frac{1}{2}(-2x)\):

\[ \begin{aligned} \int\frac{x}{5 - x^{2}}dx \\ &= -\frac{1}{2}\int\frac{-2x}{5 - x^{2}}dx \\ &= -\frac{1}{2}\log|5 - x^{2}| + C \end{aligned} \]

The option \(\frac{1}{2}\log|5 - x^{2}|\) has the wrong sign; note that \(-\frac{1}{2}\log\left|\frac{1}{5 - x^{2}}\right|\) equals that same wrong expression. The options with coefficient \(\pm 2\) do not differentiate back to the integrand.

Answer: A

The integral is \(\frac{1}{6}\log_{e}|1 + 3e^{2x}| + C\).

Multiply top and bottom by \(e^{x}\):

\[ \begin{aligned} \int\frac{e^{2x}}{1 + 3e^{2x}}dx \\ &= \frac{1}{6}\int\frac{6e^{2x}}{1 + 3e^{2x}}dx \\ &= \frac{1}{6}\log|1 + 3e^{2x}| + C \end{aligned} \]

because the derivative of \(1 + 3e^{2x}\) is \(6e^{2x}\). \(\frac{1}{3}\) forgets the 2 from \(e^{2x}\), and the options in \(e^{-x} + 3e^{2x}\) do not differentiate back to the integrand.

Answer: B

The integral is 6.

On \([-1, 1]\), \(x - 3 \lt 0\), so \(|x - 3| = 3 - x\):

\[ \begin{aligned} \int_{-1}^{1}(3 - x)\,dx \\ &= \left[3x - \frac{x^{2}}{2}\right]_{-1}^{1} \\ &= 2.5 - (-3.5) \\ &= 6 \end{aligned} \]

An integral of an absolute value cannot be negative, which rules out \(-5\) and \(-6\); 0 and 5 come from mishandling the \(x\) term.

Geometrically it is the area of a trapezium with parallel sides 4 and 2 (the values of \(3 - x\) at \(-1\) and 1) and width 2: \(\frac{4 + 2}{2} \times 2 = 6\).

Answer: E

The area is \(5\pi - 10\).

The region lies between the chord from \((0, 5)\) to \((4, 0)\) and the arc of the ellipse in the first quadrant:

\[ \begin{aligned} \text{quarter ellipse} &= \frac{\pi ab}{4} = \frac{\pi \times 4 \times 5}{4} = 5\pi \\ \text{triangle} &= \frac{1}{2}\times 4 \times 5 = 10 \end{aligned} \]

The required area is \(5\pi - 10\). \(10\pi - 5\) swaps the roles, and the options with \(4\pi\) use \(a = b = 4\).

Answer: A

The integral is 1.

Replace \(x\) by \(2 - x\), which does not change a definite integral over \([0, 2]\):

\[ I = \int_{0}^{2}\frac{(2 - x)^{4}}{(2 - x)^{4} + x^{4}}dx \]

Adding the two forms, the integrand becomes 1, so \(2I = \int_{0}^{2}1\,dx = 2\) and \(I = 1\). 2 is \(2I\), forgetting to halve, and the other values do not follow.

This 'king' property, \(\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a - x)dx\), always gives half the length of the interval for integrands of the form \(\frac{g(x)}{g(x) + g(a - x)}\).

Answer: E

The integral is 1.

\(\frac{\tan t}{\cos t} = \sec t\tan t\), whose antiderivative is \(\sec t\):

\[ \begin{aligned} \int_{0}^{\pi/3}\sec t\tan t\,dt \\ &= \sec\frac{\pi}{3} - \sec 0 \\ &= 2 - 1 \\ &= 1 \end{aligned} \]

2 forgets the lower limit, and \(\pm\frac{1}{2}\) or \(-2\) use \(\cos\) in place of \(\sec\).

The integrand is the derivative of \(\sec t\), which is the step to recognise: \(\frac{d}{dt}\sec t = \sec t\tan t\).