Integral Calculus Questions and Answers

Answer: B

Take \( u = \ln x \) and \( dv = dx \), since the logarithm has no elementary antiderivative of its own:

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

The general antiderivative is \( x\ln x - x \), which is worth remembering because the logarithm arises so often as one factor of a product

Answer: A

Write \( 1 = A(x-2) + B(x-1) \). Setting \( x = 1 \) gives \( A = -1 \), and setting \( x = 2 \) gives \( B = 1 \), so

\[ \frac{1}{(x-1)(x-2)} \;=\; \frac{-1}{x-1} + \frac{1}{x-2}. \]

A quick check at \( x = 0 \) gives \( 1/2 \) on both sides, so the signs are the right way round

Answer: D

A factor \( (x-a)^{2} \) needs \( A/(x-a) + B/(x-a)^{2} \), two constants for the two degrees of freedom lost. One term alone cannot represent a general numerator of degree one over the square. A linear numerator over a squared factor is the form reserved for an irreducible *quadratic* factor

Answer: C

The integrand is an odd function, \( (-x)^{3} = -x^{3} \), and the interval is symmetric about the origin, so the two halves cancel:

\[ \int_{-a}^{a} x^{3}dx \;=\; \left[\frac{x^{4}}{4}\right]_{-a}^{a} \;=\; \frac{a^{4}}{4} - \frac{a^{4}}{4} \;=\; 0. \]

For an even integrand the halves would add instead, giving twice the integral over half the range

Answer: B

An even function satisfies \( f(-x) = f(x) \), so the graph is a mirror image about the vertical axis and the two halves of the interval contribute equally:

\[ \int_{-a}^{a} f(x)\,dx \;=\; 2\int_{0}^{a} f(x)\,dx. \]

Vanishing is the odd case, and shifting the range to \( [0, 2a] \) samples different values of the function altogether

Answer: A

By definition \( \int_{b}^{a} f = -\int_{a}^{b} f \), which is what makes the additivity rule \( \int_{a}^{b} + \int_{b}^{c} = \int_{a}^{c} \) hold for any arrangement of the three points. The convention is not a separate rule but a consequence of writing the increment of the antiderivative as \( F(b) - F(a) \)

Answer: D

The Riemann definition partitions the interval, forms \( \sum f(x_{i})\,\Delta x_{i} \), and takes the limit as the largest width tends to zero. That the result equals \( F(b) - F(a) \) is the content of the fundamental theorem, a proved connection rather than the definition itself

Answer: C

Evaluate to a finite upper limit and then let it grow:

\[ \int_{1}^{t} \frac{dx}{x^{2}} \;=\; \left[-\frac{1}{x}\right]_{1}^{t} \;=\; 1 - \frac{1}{t} \;\longrightarrow\; 1 \quad (t \to \infty). \]

The tail dies fast enough for the total area to be finite, whereas the same integral with \( 1/x \) in place of \( 1/x^{2} \) diverges logarithmically

Answer: B

For \( p \neq 1 \) the antiderivative is \( x^{1-p}/(1-p) \), which stays finite as \( x \to \infty \) only when \( 1 - p < 0 \). At \( p = 1 \) the antiderivative is \( \ln x \), which diverges. So the boundary case itself fails, and the condition is strict: \( p > 1 \)

Answer: A

The first kind has an infinite limit of integration; the second has a finite interval but an integrand that blows up at an endpoint or an interior point, as \( \int_{0}^{1} dx/\sqrt{x} \) does at the origin. A change of sign raises no difficulty at all, and a removable discontinuity does not affect the value