Exercise 7.10 · Q9: $\int_0^2x\sqrt{2-x}\,dx$

Basics you need

Solution

1. The integrand is continuous on $[0, 2]$

   For $0 \le x \le 2$⁠, $2 - x \ge 0$⁠, so $\sqrt{2 - x}$ is defined and continuous, and so is $x\sqrt{2 - x}$⁠. Label

   $$I = \int_0^2 x\sqrt{2 - x}\,dx \qquad (1)$$
   P4 and Theorem 2 need the integrand continuous on the closed interval. The upper limit 2 is exactly where the root's inside reaches 0; beyond it the integrand would not be defined.
   $\sqrt{u}$ is defined and continuous for $u \ge 0$⁠.

2. Apply P4 with $a = 2$

   Here $f(x) = x\sqrt{2 - x}$⁠. Replace every $x$ by $2 - x$⁠:

   $$\begin{aligned}f(2 - x) &= (2 - x)\sqrt{2 - (2 - x)} \\ &= (2 - x)\sqrt{x} .\end{aligned}$$

   So, by P4,

   $$I = \int_0^2 (2 - x)\sqrt{x}\,dx \qquad (2)$$
   P4 holds for every continuous function on $[0, a]$ (Step 1). Its use: $2 - (2 - x) = x$⁠, so the root now contains $x$ alone, which the power rule can handle.
   P4, Section 7.10. Lesson: properties of definite integrals

3. Write the integrand as powers of $x$

   $$(2 - x)\sqrt{x} = 2x^{\frac12} - x \cdot x^{\frac12} = 2x^{\frac12} - x^{\frac32} .$$
   The power rule integrates one power at a time, so we expand the bracket and combine $x \cdot x^{1/2}$ into a single power.
   $x^1 \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}$ (add the indices). Basics: powers of x

4. Integrate with the power rule

   $\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac23 x^{3/2}$ and $\int x^{3/2}\,dx = \frac25 x^{5/2}$⁠. So

   $$\begin{aligned}I &= \left[\frac43 x^{\frac32} - \frac25 x^{\frac52}\right]_0^2 \\ &= \frac43 \cdot 2^{\frac32} - \frac25 \cdot 2^{\frac52} - 0 .\end{aligned}$$
   Dividing by the new index $\frac32$ is multiplying by $\frac23$⁠; with the factor 2 in front this gives $\frac43$⁠. At the lower limit both powers of 0 are 0, because their indices are positive.
   $\int x^m\,dx = \frac{x^{m+1}}{m+1} + C$⁠, with $m = \frac12$ and $m = \frac32$ (standard formula (i), Section 7.2); Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

5. Simplify the surds

   $2^{3/2} = 2\sqrt2$ and $2^{5/2} = 4\sqrt2$⁠, so

   $$\begin{aligned}I &= \frac{8\sqrt2}{3} - \frac{8\sqrt2}{5} = 8\sqrt2\cdot\frac{5 - 3}{15} \\ &= \frac{16\sqrt2}{15} \approx 1.5084945 .\end{aligned}$$
   Both terms contain $8\sqrt2$⁠, so take it out and subtract the fractions $\frac13 - \frac15 = \frac{2}{15}$⁠.
   $2^{3/2} = 2^1 \cdot 2^{1/2} = 2\sqrt2$⁠; $2^{5/2} = 2^2 \cdot 2^{1/2} = 4\sqrt2$⁠. Basics: powers of x

6. A second route, and the picture

   Without P4, write $x = 2 - (2 - x)$⁠:

   $$x\sqrt{2 - x} = 2(2 - x)^{\frac12} - (2 - x)^{\frac32} .$$

   An antiderivative is $F(x) = -\frac43(2 - x)^{3/2} + \frac25(2 - x)^{5/2}$⁠. Check by the chain rule: $F'(x) = 2(2 - x)^{1/2} - (2 - x)^{3/2}$⁠, the integrand. Then

   $$\begin{aligned}I &= F(2) - F(0) = 0 - \left(-\tfrac{8\sqrt2}{3} + \tfrac{8\sqrt2}{5}\right) \\ &= \frac{16\sqrt2}{15} .\end{aligned}$$
   The second route does by hand what P4 does in one line (it amounts to the substitution $t = 2 - x$⁠). Both give the same value, which confirms the answer.
   Chain rule: $\frac{d}{dx}(2 - x)^{k} = -k(2 - x)^{k - 1}$⁠. Basics: substitution

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