Definite and Indefinite Integrals Questions and Answers

Answer: D

The integral is \(\frac{x+1}{2}\sqrt{x^{2}+2x+3} + \log\left|(x+1) + \sqrt{x^{2}+2x+3}\right| + C\).

Complete the square: \(x^{2} + 2x + 3 = (x + 1)^{2} + (\sqrt{2})^{2}\). The standard result with \(u = x + 1\) and \(a^{2} = 2\) is:

\[ \begin{aligned} \int\sqrt{u^{2} + a^{2}}\,du &= \frac{u}{2}\sqrt{u^{2} + a^{2}} \\ &\quad + \frac{a^{2}}{2}\log\left|u + \sqrt{u^{2} + a^{2}}\right| \\ &\quad + C \end{aligned} \]

Here \(\frac{a^{2}}{2} = 1\), so the logarithm has coefficient 1 and a plus sign, as in the fourth option.

The first option forgets the \(\frac{1}{2}\) in front of the first term. The options with a minus before the log, or with \((x+1) - \sqrt{\cdots}\) inside it, do not differentiate back to \(\sqrt{x^{2} + 2x + 3}\).

Answer: B

The integral equals 1.

Use \(\int_{a}^{b}f(x)\,dx = \int_{a}^{b}f(a + b - x)\,dx\). Here \(a + b = 8\), and replacing \(x\) by \(8 - x\) swaps the two exponentials:

\[ \begin{aligned} I &= \int_{3}^{5}\frac{e^{(1+x^{2})}\,dx}{e^{(1+x^{2})} + e^{(1+(8-x)^{2})}} \\ I &= \int_{3}^{5}\frac{e^{(1+(8-x)^{2})}\,dx}{e^{(1+(8-x)^{2})} + e^{(1+x^{2})}} \\ 2I &= \int_{3}^{5}1\,dx = 2, \quad I = 1 \end{aligned} \]

Adding the two forms makes the integrand 1. The answer is half the interval length, so 2 forgets to halve, and 5 or 3 are the limits themselves.

Answer: D

The integral equals \(6a\).

Over a symmetric interval \([-a, a]\), an odd function integrates to 0. Check each term:

• \(x^{3}\) is odd.
• \(x\cos^{2}2x\) is odd, since \(\cos^{2}2x\) is even and \(x\) is odd.
• \(\tan^{3}x\) is odd.
• 3 is even.

\[ \int_{-a}^{a}3\,dx = 3(a - (-a)) = 6a \]

3a integrates the constant over \([0, a]\) only, and 2a and 4a do not come from any of the terms. Only the constant survives.

Answer: E

The area is \(\frac{9}{2}\) sq. units.

The curve meets the \(x\)-axis where \(3x - x^{2} = x(3 - x) = 0\), at \(x = 0\) and \(x = 3\), and lies above the axis in between:

\[ \begin{aligned} A &= \int_{0}^{3}(3x - x^{2})\,dx \\ &= \left[\frac{3x^{2}}{2} - \frac{x^{3}}{3}\right]_{0}^{3} \\ &= \frac{27}{2} - 9 = \frac{9}{2} \end{aligned} \]

\(\frac{27}{2}\) is the first term alone, forgetting to subtract \(\frac{x^{3}}{3}\). 9 is the subtracted term, and 18 doubles it.

Answer: C

The area is 8 sq.units.

For \(x \ge 0\), \(y = |x|\) is the line \(y = x\). Together with the line \(x = 4\) and the \(x\)-axis it closes a right triangle with vertices \((0, 0)\), \((4, 0)\) and \((4, 4)\), which is the region the key takes:

\[ A = \int_{0}^{4}x\,dx = \frac{1}{2}(4)(4) = 8 \]

The \(x\)-axis is the implied third boundary, as in the usual 'area under the curve' reading; the other branch \(y = -x\) lies at \(x \le 0\), on the other side of the origin from the line \(x = 4\). 4 halves the triangle again, and 12 and 13 do not correspond to any region bounded by these lines.

Answer: B

\(f(x) = 2x^{3} + 3x^{2} - 5x - 7\).

Integrate \(\frac{dy}{dx} = 6x^{2} + kx - 5\):

\[ f(x) = 2x^{3} + \tfrac{k}{2}x^{2} - 5x + c \]

Use the two points:

\[ \begin{aligned} f(1) &= 2 + \tfrac{k}{2} - 5 + c = -7 \\ f(2) &= 16 + 2k - 10 + c = 11 \end{aligned} \]

These give \(\frac{k}{2} + c = -4\) and \(2k + c = 5\), so \(k = 6\) and \(c = -7\). The \(x^{2}\) coefficient is \(\frac{k}{2} = 3\). Check: \(f(2) = 16 + 12 - 10 - 7 = 11\). The options with \(6x^{2}\) forget to halve \(k\) when integrating.

Answer: A

The integral is \(x(\sin x - \cos x) + \cos x + \sin x + C\).

Since \(\cos 2x = (\cos x - \sin x)(\cos x + \sin x)\), the integrand simplifies to \(x(\cos x + \sin x)\). Integrate by parts with \(u = x\):

\[ \begin{aligned} \int x(\cos x + \sin x)dx &= x(\sin x - \cos x) \\ &\quad - \int(\sin x - \cos x)dx \\ &= x(\sin x - \cos x) + \cos x + \sin x + C \end{aligned} \]

Check by differentiating: the derivative is \((\sin x - \cos x) + x(\cos x + \sin x) - \sin x + \cos x = x(\cos x + \sin x)\). The other options fail this check.

Answer: E

The integral is \(\log_{e}|\sin x| + C\).

Combine the two terms with the double-angle formulas:

\[ \begin{aligned} \cot 2x + \mathrm{cosec}\,2x &= \frac{\cos 2x + 1}{\sin 2x} \\ &= \frac{2\cos^{2}x}{2\sin x\cos x} = \cot x \end{aligned} \]

\[ \int \cot x\,dx = \log_{e}|\sin x| + C \]

Integrating each term separately and then forgetting to simplify can suggest a factor of \(\frac{1}{2}\) or 2, but the combined integrand is exactly \(\cot x\).

Answer: C

\(a = -10\).

Split the integrand into powers of \(x\):

\[ \frac{5\sqrt{x}(4x^{2} - 1)}{x} = 20x^{3/2} - 5x^{-1/2} \]

\[ \begin{aligned} \int (20x^{3/2} - 5x^{-1/2})dx &= 20 \cdot \tfrac{2}{5}x^{5/2} - 5 \cdot 2x^{1/2} \\ &= 8\sqrt{x^{5}} - 10\sqrt{x} + C \end{aligned} \]

The first term matches the given \(8\sqrt{x^{5}}\), confirming the working, and the coefficient of \(\sqrt{x}\) is \(-10\). The option 10 has the sign of the second term reversed.

Answer: C

The integral is \(x + \frac{8}{x + 3} + C\).

Rewrite the numerator around \((x + 3)^{2} = x^{2} + 6x + 9\):

\[ \begin{aligned} \frac{x^{2} + 6x + 1}{(x + 3)^{2}} &= \frac{(x + 3)^{2} - 8}{(x + 3)^{2}} \\ &= 1 - \frac{8}{(x + 3)^{2}} \end{aligned} \]

\[ \begin{aligned} &\int\left(1 - \frac{8}{(x + 3)^{2}}\right)dx \\ &= x + \frac{8}{x + 3} + C \end{aligned} \]

Because \(\int (x + 3)^{-2}dx = -(x + 3)^{-1}\), the minus signs combine to give \(+\frac{8}{x + 3}\). The option \(x - \frac{8}{x + 3}\) misses this sign change.