# Recitation 11

Problem 1 Prove the reduction formula for $n\neq 1$: $$\int\tan^n x dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2} x dx.$$

Idea Rewrite $\tan^{n}x = \tan^{n-2}x(\sec^2 x - 1)$ and use $(\tan x)' = \sec^2x$.

Fact 1 If $g(t) = \int_{a(t)}^{b(t)} f(x) dx$, then $g'(t) = f(a(t))a'(t) - f(b(t))b'(t)$. Prove it by the fundamental theorem of calculus.

Problem 2 Consider the function $f\colon [0,1] \to \mathbb{R}$ defined by $f(x) = e^{x^2}$. For every $0 < t < 1$ let $A(t)$ denote the area below the graph of $f$ and above the x-axis for $0. Let $B(t)$be the area above the graph of $f$ and below the line $y = e$ for $t < x < 1$. Find the maximum point of $A(t)+B(t)$.

Solution Note that $A(t) = \int_0^t f(x) dx$ and $B(t) = \int_t^1 (e - f(x)) dx$. Let $S(t) = A(t) + B(t)$. By the fundamental theorem of calculus, $$S'(t) = A'(t) + B'(t) = f(t) - (e - f(t)) = 2f(t) - e.$$ Solve $S'(t) = 0$ and get $t^* = \sqrt{1-\ln 2}$. When $0 \le t < t^*$, $S'(t) < 0$ and so $S$ is monotone decreasing; when $t^* < t \le 1$, $S'(t) > 0$ and so $S$ is monotone increasing. Therefore $t = 0, 1$ are local maximum points. We only need to compare $S(0)$ and $S(1)$. Because $S(0) + S(1) = e$, it is enough to compare $S(1)$ with $e/2$. Notice that $S(1) = \int_0^1 f(x) dx$ and we can approximate the integral from below by $$(f(0) + f(1/8) + f(2/8) + f(3/8) + f(4/8) + f(5/8) + f(6/8) + f(7/8))/8 = 1.36... > e/2.$$ Therefore $S(1) > e/2 > S(0)$ and $t=1$ is the maximum point.

Alternative solution The last part to show $\int_0^1 e^{x^2}dx > e/2$ can be simplified as follows. Because $0 \le x \le 1$ and $e^{x^2} \ge 1$, $e^{x^2} - 1 \ge (e^{x^2} - 1)x$ and so $e^{x^2} \ge \left(e^{x^2} - 1\right)x + 1$. In other words, the graph of $e^{x^2}$ is above the graph of $\left(e^{x^2} - 1\right)x + 1$ for $0 \le x \le 1$. Therefore $$\int_0^1 e^{x^2} dx > \int_0^1 \left[\left(e^{x^2} - 1\right)x + 1\right] dx = \left.\frac{1}{2}e^{x^2} - \frac{1}{2}x^2 + x \right|_0^1 = e/2.$$

Problem 3 Find the derivative of $F(x) = \int_{0}^{x^2}|\sin t|dt$.

Answer $F'(x) = |\sin x^2| 2x$.

Problem 4 Find the limit $\lim_{x\to 0}\frac{\int_0^x (1-\cos t)dt}{x^2\sin x}dt$.

Answer We know that $\frac{\int_0^x (1-\cos t)dt}{x^2\sin x} = \frac{\int_0^x (1-\cos t)dt}{x^3} \frac{x}{\sin x}$. It suffices to compute $\lim_{x\to 0}\frac{\int_0^x (1-\cos t)dt}{x^3}$. This limit is of type 0/0. We consider $$\lim_{x\to 0}\frac{\left(\int_0^x (1-\cos t)dt\right)'}{\left(x^3\right)'} = \lim_{x\to 0}\frac{1-\cos x}{3x^2},$$ which is again a limit of type 0/0. We consider $$\lim_{x\to 0}\frac{\left(1-\cos x\right)'}{\left(3x^2\right)'} = \lim_{x\to 0}\frac{\sin x}{6x} = \frac{1}{6}.$$ By l’Hôpital’s rule twice, we get the limit is 1/6.

Problem 5 Consider the one-to-one and onto function $f\colon [0,2] \to [1,6]$ defined by $f(x) = x + 2^x$. Show that $\int_0^2 f(x) dx + \int_1^6 f^{-1}(x) dx = 12$.

Geometric solution Observe that $\int_0^2 f(x) dx$ is the area bounded by $y = f(x)$, $y = 0$ and $x = 2$, and $\int_1^6 f^{-1}(x) dx$ is the area bounded by $y = f(x)$, $x = 0$ and $y = 6$. Thus the sum is the area of the rectangle bounded by $x = 0, 2$ and $y = 1,6$.

Algebraic solution For the second integral $\int_1^6 f^{-1}(x) dx$ consider substitution $y = f^{-1}(x)$. Thus $x = f(y), dx = f'(y) dy$ and $\int_1^6 f^{-1}(x) dx = \int_0^2 yf'(y)dy$. Using integration by parts, $\int_0^2 yf'(y)dy = \left.yf(y)\right|_0^2 - \int_0^2 f(y)dy$. Thus $$\int_0^2 f(x) dx + \int_1^6 f^{-1}(x) dx = \int_0^2 f(y)dy + \int_0^2 yf'(y)dy = \left.yf(y)\right|_0^2 = 12.$$

Problem 6 Find the area bounded by the arcs of the parabolas $y = x^2 / 4$ and $y = 3 - x^2 / 2$.

Idea The intersections of the parabolas are given by $x^2 / 4 = 3 - x^2 / 2$, that is $x = \pm 2$. The area is thus $\int_{-2}^2 3 - x^2 /2 - x^2 / 4$.

Problem 7 Compute the integral $\int_4^9 \frac{\sqrt{x}}{\sqrt{x} - 1}dx$.

Idea The anti-derivative of $\frac{\sqrt{x}}{\sqrt{x} - 1}$ is $x + 2\sqrt{x} + 2\ln(\sqrt{x}-1)$. By Newton–Leibniz formula, the definite integral equal $\left.x + 2\sqrt{x} + 2\ln(\sqrt{x}-1)\right|_4^9 = 7+\ln 4$.

Problem 8 Find the length of the graph of $f(x) = x^2$ for $0\le x \le 1$.

Solution The length is given by $$\int_0^1 \sqrt{1 + f'(x)^2} dx = \int_0^1 \sqrt{1 + 4x^2} dx.$$ The integral of $\int_0^1 \sqrt{1 + 4x^2} dx$ can be obtained by Euler substitution $\sqrt{1 + 4x^2} = t + 2x$. We get $x = \frac{1-t^2}{4t}$, $t+2x = \frac{1+t^2}{2t}$ and $dx = \frac{1+t^2}{4t^2}dt$. Thus $$\int \sqrt{1 + 4x^2} dx = \int \frac{(1+t^2)^2}{8t^3}dt = \frac{t^4 + 4t^2\ln t - 1}{16t^2}.$$ On the other hand, by our substitution $t = \sqrt{1 + 4x^2} - 2x$. Plug this in and get $$\int \sqrt{1 + 4x^2} dx = \frac{1}{2}x\sqrt{1+4x^2} + \frac{1}{4}\ln (\sqrt{1 + 4x^2} - 2x).$$ By Newton–Leibniz formula, the definite integral equals $\frac{1}{2}\sqrt{5} + \frac{1}{4}\ln (\sqrt{5}-2)$.

Alternative solution We can also use trigonometric substitution $x = \frac{1}{2}\tan u$ for $\int \sqrt{1 + 4x^2} dx$ and use reduction formula for $\int 1/\cos^3 u du$.

Problem 9 Prove that for any function $f\colon [0,1]\to \mathbb{R}$ $$\int_0^\pi xf(\sin x) dx = \frac{\pi}{2}\int_0^\pi f(\sin x)dx.$$

Algebraic solution Use substitution $x = \pi - u$ and get $$\int_0^\pi xf(\sin x) dx = \int_\pi^0 (\pi - u)f(\sin u)(-du) = \int_0^\pi (\pi - u)f(\sin u)du.$$ Therefore $$2\int_0^\pi xf(\sin x) dx = \int_0^\pi xf(\sin x) dx + \int_0^\pi (\pi - u)f(\sin u)du = \int_0^\pi \pi f(\sin x) dx.$$

Question Can you interpret the algebraic solution geometrically?

Problem 10 Let $A$ be the area bounded between the graph of $f (x) = \sin x$ and the graph of $g(x)=\cos x$ where $\pi/4 \le x \le 2\pi +\pi /4$. Find the volume of the solid obtained by revolving $A$ about the horizontal line $y = 4$.

Idea Notice that the graphs of $\sin x$ is above the graph of $\cos x$ for $\pi / 4 \le x \le \pi + \pi/4$ and the other way around for $\pi+\pi / 4 \le x \le 2\pi + \pi/4$. The volume is thus given by $$\int_{\pi/4}^{\pi + \pi / 4} \pi\left((4 - \cos x)^2 - (4 - \sin x)^2\right) dx + \int_{\pi + \pi/4}^{2\pi + \pi / 4} \pi\left((4 - \sin x)^2 - (4 - \cos x)^2\right) dx.$$