Integral Calculus Questions and Answers

Answer: A

The enclosed area is 1/6.

The line and the parabola meet where \(x = x^{2}\), at x = 0 and x = 1. Between these points the line lies above the parabola, so:

\[ \begin{aligned} A &= \int_{0}^{1}(x-x^{2})\,dx \\ &= \left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} \\ &= \frac{1}{2}-\frac{1}{3} = \frac{1}{6} \end{aligned} \]

1/2 is the area under the line alone and 1/3 the area under the parabola alone; the region between them is their difference. 1/4 does not correspond to either.

Answer: C

The limit is 5.

Let \(F(x) = \int_{2}^{2+x} g(t)\,dt\) with \(g(t) = t + \sqrt{t^{2} + 5}\). Then \(F(0) = 0\), so the limit is the derivative of F at 0, which by the fundamental theorem of calculus is the integrand at the upper limit:

\[ \begin{aligned} \lim_{x \to 0}\frac{F(x)}{x} &= F'(0) = g(2) \\ &= 2 + \sqrt{4 + 5} = 2 + 3 = 5 \end{aligned} \]

0 treats the integral as vanishing faster than x, but over an interval of length x the integral is about x times the integrand, which is why the ratio tends to g(2). 4 and 6 are not the value of the integrand at t = 2.

Answer: C

\(f(2) = 2 + \log_{e} 4\).

Multiply the defining relation by x and differentiate, using \(g'(x) = f(x)\):

\[ \begin{aligned} x f(x) &= 2x + g(x) \\ f + x f' &= 2 + f \\ f'(x) &= \frac{2}{x} \end{aligned} \]

So \(f(x) = 2\ln x + C\). At x = 1, \(g(1) = 0\), so \(f(1) = 2\) and C = 2. Hence \(f(2) = 2 + 2\ln 2 = 2 + \log_{e} 4\).

\(2 + \log_{e} 2\) drops the factor 2 from \(f' = 2/x\), and the two options with a minus sign reverse the sign of the logarithm.

Answer: A

Raising the index and dividing by the new index reverses the power rule:

\[ \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) \;=\; \frac{(n+1)x^{n}}{n+1} \;=\; x^{n}. \]

The exclusion \( n = -1 \) is needed because the divisor would vanish; that case integrates to \( \ln|x| \) instead. Lowering the index is differentiation, not integration

Answer: D

Integrate and evaluate at the two limits:

\[ \int_{0}^{\pi/2} \sin x\,dx \;=\; \left[-\cos x\right]_{0}^{\pi/2} \;=\; -\cos\frac{\pi}{2} + \cos 0 \;=\; 0 + 1 \;=\; 1. \]

The value two belongs to the integral over the full half-period from zero to \( \pi \), which covers twice the area of this quarter-period

Answer: C

The reciprocal integrates to the natural logarithm:

\[ \int_{1}^{e} \frac{dx}{x} \;=\; \left[\ln x\right]_{1}^{e} \;=\; \ln e - \ln 1 \;=\; 1 - 0 \;=\; 1. \]

This is the defining property of \( e \): it is the upper limit at which the area under the reciprocal curve, measured from one, first reaches unity

Answer: B

Substituting \( x = a\tan\theta \) gives \( dx = a\sec^{2}\theta\,d\theta \) and turns the denominator into \( a^{2}\sec^{2}\theta \), leaving

\[ \int \frac{a\sec^{2}\theta\,d\theta}{a^{2}\sec^{2}\theta} \;=\; \frac{\theta}{a} \;=\; \frac{1}{a}\tan^{-1}\frac{x}{a} + C. \]

The inverse sine belongs to \( 1/\sqrt{a^{2}-x^{2}} \), and the logarithm to \( 1/(a^{2}-x^{2}) \)

Answer: A

Integrating \( (uv)' = u'v + uv' \) and rearranging gives \( \int u\,dv = uv - \int v\,du \). So the method undoes a product rule. The chain rule's counterpart is substitution, and the quotient rule has no integral method of its own, quotients being handled by partial fractions or substitution instead

Answer: D

Take \( u = x \) and \( dv = e^{x}dx \), so that differentiating the polynomial factor removes it:

\[ \int_{0}^{1} x e^{x}dx \;=\; \left[x e^{x}\right]_{0}^{1} - \int_{0}^{1} e^{x}dx \;=\; e - \left[e^{x}\right]_{0}^{1} \;=\; e - (e - 1) \;=\; 1. \]

Choosing the factors the other way round would raise the power of \( x \) and make the integral harder rather than easier

Answer: C

ILATE ranks inverse trigonometric, logarithmic, algebraic, trigonometric and exponential, and the earlier of the two factors is taken as \( u \). The point is that inverse and logarithmic functions simplify sharply on differentiating but are awkward to integrate, so they belong in the part that gets differentiated