Integral Calculus Questions and Answers

Answer: C

Integrate term by term and substitute the limits:

\[ \int_{0}^{2}\left(3x^{2} + 2x\right)dx \;=\; \left[x^{3} + x^{2}\right]_{0}^{2} \;=\; 8 + 4 \;=\; 12. \]

The two pieces contribute eight and four respectively, so taking either alone accounts for the eight offered as a distractor

Answer: B

The reciprocal integrates to a logarithm, and the difference of logarithms is the logarithm of the ratio:

\[ \int_{1}^{4}\frac{dx}{x} \;=\; \left[\ln x\right]_{1}^{4} \;=\; \ln 4 - \ln 1 \;=\; \ln 4 \;\approx\; 1.386. \]

The value \( \ln 3 \) would arise from taking the difference of the limits inside the logarithm rather than their ratio

Answer: A

Here \( y' = 3 \), a constant, so the integrand is constant too:

\[ L \;=\; \int_{0}^{4}\sqrt{1 + 3^{2}}\,dx \;=\; \sqrt{10}\,\left[x\right]_{0}^{4} \;=\; 4\sqrt{10} \;\approx\; 12.65. \]

Pythagoras on the horizontal run of four and the vertical rise of twelve gives the same \( \sqrt{16 + 144} = \sqrt{160} \), confirming it

Answer: D

With \( y^{2} = x \) the disc formula becomes an integral of \( x \) itself:

\[ V \;=\; \pi\int_{0}^{4} x\,dx \;=\; \pi\left[\frac{x^{2}}{2}\right]_{0}^{4} \;=\; 8\pi. \]

Forgetting to square the radius would leave \( \pi\int\sqrt{x}\,dx = 16\pi/3 \) instead, so the squaring is what makes the integrand elementary here

Answer: C

Divide the integral by the length of the interval:

\[ \bar{f} \;=\; \frac{1}{3}\int_{0}^{3} x^{2}dx \;=\; \frac{1}{3}\left[\frac{x^{3}}{3}\right]_{0}^{3} \;=\; \frac{1}{3}(9) \;=\; 3. \]

The mean exceeds the midpoint value \( f(1.5) = 2.25 \), as it must for a convex function, and falls well below the maximum of nine at the right-hand end

Answer: B

With \( h = 0.5 \) the ordinates are \( 1 \), \( 1/1.5 = 0.6667 \) and \( 0.5 \), so

\[ I \;\approx\; \frac{0.5}{3}\left[1 + 4(0.6667) + 0.5\right] \;=\; \frac{0.5}{3}(4.1667) \;=\; 0.6944. \]

The exact value is \( \ln 2 = 0.6931 \), so two strips already give three-figure accuracy, which is why the rule is preferred to the trapezoidal one

Answer: A

Arc length integrates the arc element alone. A volume slice is a disc of area \( \pi y^{2} \) and thickness \( dx \). A surface band has circumference \( 2\pi y \) and width equal to the arc element, so it carries both factors, which is what distinguishes it from the volume

Answer: D

The sketch settles which variable runs and between what values, so it comes first. The axis then determines what counts as the radius of a slice, which is squared and scaled to give the slice volume. Integration is the last step, and attempting it before the radius is identified is what produces the wrong power of \( y \)

Answer: C

The constant cancels in \( F(b) - F(a) \), so it cannot affect the value: this is precisely why a definite integral is well defined although the antiderivative is not unique. Additivity, being a number, and the sign change on exchanging limits are all genuine properties

Answer: B

With \( x = a\sin\theta \) the radical becomes \( a\cos\theta \), removing the root cleanly. The tangent substitution suits \( a^{2} + x^{2} \), and the secant and hyperbolic cosine suit \( x^{2} - a^{2} \); each of those three leaves the wrong sign under the root here and produces an imaginary result