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.