Example 25: Four definite integrals by the second fundamental theorem

Basics you need

Part (i): $\int_2^3 x^2\,dx$

1. Continuity and an antiderivative

   $f(x) = x^2$ is a polynomial, so it is continuous on $[2, 3]$⁠. By the power rule with $n = 2$⁠,

   $$\int x^2\,dx = \frac{x^3}{3} + C, \quad\text{so take}\quad F(x) = \frac{x^3}{3}.$$

   Check: $F'(x) = \frac{3x^2}{3} = x^2$⁠.

   Theorem 2 needs continuity on the closed interval and any one antiderivative; both are immediate for a polynomial.
   Standard formula (i), Section 7.2, with $n = 2$⁠. Basics: the power rule

2. Evaluate $F(3) - F(2)$

   $$I = F(3) - F(2) = \frac{27}{3} - \frac{8}{3} = \frac{19}{3} \approx 6.3333333.$$
   The second fundamental theorem turns an area into a subtraction. The figure gives a sanity check: the curve rises from 4 to 9, so the area over a base of width 1 must be between 4 and 9.
   $\big[F(x)\big]_a^b$ is shorthand for $F(b) - F(a)$⁠: upper limit first. Lesson: the fundamental theorem

Part (ii): $\int_4^9 \frac{\sqrt{x}}{(30 - x^{3/2})^2}\,dx$

1. Check continuity on $[4, 9]$

   $\sqrt{x}$ and $x^{\frac32} = x\sqrt{x}$ are continuous for $x \ge 0$⁠. As $x$ goes from 4 to 9, $x^{\frac32}$ increases from $4^{\frac32} = 8$ to $9^{\frac32} = 27$⁠. So

   $$3 \le 30 - x^{\frac32} \le 22 \quad\text{on } [4, 9].$$

   The denominator $\left(30 - x^{\frac32}\right)^2$ is therefore at least 9 and never 0, and the integrand is continuous on $[4, 9]$⁠.

   A quotient of continuous functions is continuous wherever the denominator is not 0. The denominator would vanish only where $x^{\frac32} = 30$⁠, at $x = 30^{\frac23} \approx 9.65$⁠, which is outside $[4, 9]$⁠.
   $4^{\frac32} = (\sqrt4)^3 = 2^3 = 8$ and $9^{\frac32} = (\sqrt9)^3 = 27$⁠. Basics: powers of x

2. Substitute $t = 30 - x^{\frac32}$

   Differentiate: $\frac{dt}{dx} = -\frac32 x^{\frac12} = -\frac32\sqrt{x}$⁠. So

   $$dt = -\tfrac32\sqrt{x}\,dx, \qquad \sqrt{x}\,dx = -\tfrac23\,dt.$$

   The numerator $\sqrt{x}\,dx$ is exactly what appears in $dt$⁠, so

   $$\int\frac{\sqrt{x}\,dx}{\left(30 - x^{\frac32}\right)^2} = -\frac23\int\frac{dt}{t^2} = -\frac23\int t^{-2}\,dt.$$
   The derivative of the inside of the bracket, $30 - x^{\frac32}$⁠, is a constant times $\sqrt x$⁠, and $\sqrt x$ is sitting in the numerator. That is the signal for this substitution: everything in $x$ disappears.
   Power rule for derivatives: $\frac{d}{dx}x^{\frac32} = \frac32 x^{\frac12}$⁠. Basics: substitution

3. Integrate and return to $x$

   By the power rule with $n = -2$⁠, $\int t^{-2}\,dt = \frac{t^{-1}}{-1} = -\frac{1}{t}$⁠. Hence

   $$-\frac23\left(-\frac1t\right) = \frac{2}{3t} = \frac{2}{3\left(30 - x^{\frac32}\right)} = F(x).$$

   Check: $F(x) = \frac23\left(30 - x^{\frac32}\right)^{-1}$⁠, so by the chain rule

   $$\begin{aligned}F'(x) &= \frac23\cdot(-1)\left(30 - x^{\frac32}\right)^{-2}\cdot\left(-\frac32\sqrt{x}\right) \\ &= \frac{\sqrt{x}}{\left(30 - x^{\frac32}\right)^2}.\end{aligned}$$
   Two minus signs meet here: one from $dt = -\frac32\sqrt x\,dx$ and one from $\int t^{-2}dt = -t^{-1}$⁠. They cancel, so $F$ is positive. The book's line "$= \frac23\left[\frac1t\right]$" is this same cancellation.
   Standard formula (i), Section 7.2, with $n = -2$ (allowed, since only $n = -1$ is excluded). Basics: the power rule

4. Evaluate $F(9) - F(4)$

   $$\begin{aligned}I &= \frac23\left[\frac{1}{30 - 27} - \frac{1}{30 - 8}\right] \\ &= \frac23\left[\frac13 - \frac{1}{22}\right] = \frac23\cdot\frac{22 - 3}{66} \\ &= \frac23\cdot\frac{19}{66} = \frac{19}{99} \approx 0.1919192.\end{aligned}$$

   Same answer with new limits. When $x = 4$⁠, $t = 22$⁠; when $x = 9$⁠, $t = 3$⁠. Then

   $$\begin{aligned}I &= -\frac23\int_{22}^{3}\frac{dt}{t^2} = -\frac23\left[-\frac1t\right]_{22}^{3} \\ &= \frac23\left(\frac13 - \frac1{22}\right) = \frac{19}{99}.\end{aligned}$$
   The new limits can be used because $t = 30 - x^{\frac32}$ is differentiable and decreasing on $[4, 9]$⁠, so as $x$ runs from 4 to 9, $t$ runs once from 22 down to 3 (Section 7.9). The upper limit 3 is smaller than the lower limit 22; that is fine.
   Least common denominator: $\frac13 - \frac1{22} = \frac{22}{66} - \frac{3}{66} = \frac{19}{66}$⁠. Lesson: integration by substitution

Part (iii): $\int_1^2 \frac{x}{(x+1)(x+2)}\,dx$

1. Continuity and partial fractions

   The denominator vanishes only at $x = -1$ and $x = -2$⁠, both outside $[1, 2]$⁠, so the integrand is continuous there. The degree of the numerator (1) is less than that of the denominator (2), so it is a proper fraction of form 1 in Table 7.2:

   $$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}.$$

   Multiply by $(x+1)(x+2)$⁠: $x = A(x+2) + B(x+1)$ for every $x$⁠.

   Put $x = -1$⁠: $-1 = A(1)$⁠, so $A = -1$⁠. Put $x = -2$⁠: $-2 = B(-1)$⁠, so $B = 2$⁠.

   $$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}.$$
   Each simple fraction $\frac{1}{x + k}$ has a known integral, $\log|x + k|$⁠; the product in the denominator does not. Putting $x = -1$ kills the B term and $x = -2$ kills the A term, so each constant appears on its own. Check: $-(x+2) + 2(x+1) = x$⁠.
   Partial fractions, Section 7.5, Table 7.2 form 1. Lesson: partial fractions

2. An antiderivative

   $$\int\frac{x\,dx}{(x+1)(x+2)} = -\log|x+1| + 2\log|x+2| = F(x).$$

   Check:

   $$\begin{aligned}F'(x) &= -\frac{1}{x+1} + \frac{2}{x+2} \\ &= \frac{-(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{x}{(x+1)(x+2)}.\end{aligned}$$
   $\int\frac{dx}{x+k} = \log|x+k|$ by the shift $u = x + k$ and standard formula (xii). On $[1, 2]$ both $x + 1$ and $x + 2$ are positive, so the bars change nothing here.
   "log" in this book is the natural logarithm, base e. Basics: logarithms and 1/x

3. Evaluate $F(2) - F(1)$

   $$\begin{aligned}I &= \left[-\log 3 + 2\log 4\right] - \left[-\log 2 + 2\log 3\right] \\ &= -3\log 3 + \log 2 + 2\log 4 \\ &= \log\frac{2\cdot 4^2}{3^3} = \log\frac{32}{27} \approx 0.1698990.\end{aligned}$$
   Laws of logs combine the answer into one log: $k\log a = \log a^k$⁠, $\log a + \log b = \log ab$⁠, $\log a - \log b = \log\frac{a}{b}$⁠. The figure agrees: the curve is nearly flat at height about 0.17 over a base of width 1.
   $2\log 4 = \log 16$⁠, $3\log 3 = \log 27$⁠, so the total is $\log 2 + \log 16 - \log 27$⁠. Basics: logarithms

Part (iv): $\int_0^{\pi/4} \sin^3 2t\,\cos 2t\,dt$

1. Continuity and the substitution $u = \sin 2t$

   The integrand is built from sines and cosines, so it is continuous everywhere, in particular on $\left[0, \frac{\pi}{4}\right]$⁠. Put $u = \sin 2t$⁠. Then by the chain rule $\frac{du}{dt} = 2\cos 2t$⁠, so

   $$du = 2\cos 2t\,dt, \qquad \cos 2t\,dt = \tfrac12\,du.$$ $$\int\sin^3 2t\,\cos 2t\,dt = \frac12\int u^3\,du.$$
   $\cos 2t$ is (up to the factor 2) the derivative of $\sin 2t$⁠, and the rest of the integrand is a power of $\sin 2t$⁠. So $u = \sin 2t$ turns everything into a plain power of $u$⁠.
   $\frac{d}{dt}\sin 2t = 2\cos 2t$ (chain rule, inner function $2t$⁠). Basics: substitution

2. Integrate and return to $t$

   $$\frac12\int u^3\,du = \frac12\cdot\frac{u^4}{4} = \frac18 u^4 = \frac18\sin^4 2t = F(t).$$

   Check: $F'(t) = \frac18\cdot 4\sin^3 2t\cdot\cos 2t\cdot 2$⁠, which is $\sin^3 2t\,\cos 2t$⁠.

   The chain rule is used twice in the check: once for the fourth power and once for the inner $2t$⁠. The factors $\frac18\cdot 4\cdot 2 = 1$ are exactly what makes the answer right.
   Power rule with $n = 3$⁠. Basics: the power rule

3. Evaluate $F\left(\frac{\pi}{4}\right) - F(0)$

   $$I = \frac18\left[\sin^4\frac{\pi}{2} - \sin^4 0\right] = \frac18(1 - 0) = \frac18 = 0.125.$$

   With new limits instead: $t = 0$ gives $u = 0$⁠, $t = \frac{\pi}{4}$ gives $u = \sin\frac{\pi}{2} = 1$⁠, and $\frac12\int_0^1 u^3\,du = \frac12\cdot\frac14 = \frac18$⁠.

   At $t = \frac{\pi}{4}$⁠, $2t = \frac{\pi}{2}$ and $\sin\frac{\pi}{2} = 1$⁠. The substitution $u = \sin 2t$ is increasing on $\left[0, \frac{\pi}{4}\right]$ (since $2t \in \left[0, \frac{\pi}{2}\right]$⁠), so $u$ runs once from 0 to 1 and the new limits are legitimate.
   $\sin 0 = 0$⁠, $\sin\frac{\pi}{2} = 1$⁠. Basics: trigonometric identities

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