Chapter 7 · Integrals

Exercise 7.9 · Q4: $\int_0^2 x\sqrt{x+2}\,dx$

Exercise 7.9Definite integrals by substitution · Section 7.9

The question

Evaluate the integral using substitution:

\int_0^2 x\sqrt{x + 2}\;dx \qquad (\text{Put } x + 2 = t^2).

(The book prints the integral without the $dx$ ⁠.)

Exam answer

Let $I = \displaystyle\int_0^2 x\sqrt{x + 2}\,dx$ ⁠.

Put $x + 2 = t^2$ ( $t \ge 0$ ⁠). Then $dx = 2t\,dt$ and $x = t^2 - 2$ ⁠.

When $x = 0$ ⁠, $t = \sqrt2$ ⁠; when $x = 2$ ⁠, $t = 2$ ⁠.

\begin{aligned} \therefore\ I &= \int_{\sqrt2}^2 (t^2 - 2)\,t\cdot 2t\,dt \\ &= \int_{\sqrt2}^2\left(2t^4 - 4t^2\right)dt \\ &= \left[\frac{2t^5}{5} - \frac{4t^3}{3}\right]_{\sqrt2}^2 \\ &= \left(\frac{64}{5} - \frac{32}{3}\right) \\ &\quad - \left(\frac{8\sqrt2}{5} - \frac{8\sqrt2}{3}\right) \\ &= \frac{32}{15} + \frac{16\sqrt2}{15} \end{aligned}

$\therefore\ I = \dfrac{16\sqrt2}{15}\left(\sqrt2 + 1\right)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Naming the square root $t$ removes it: with $x + 2 = t^2$ and $t \ge 0$ ⁠, $\sqrt{x+2} = t$ ⁠, $x = t^2 - 2$ and $dx = 2t\,dt$ ⁠. The integrand becomes the polynomial $(t^2 - 2)\cdot t\cdot 2t = 2t^4 - 4t^2$ ⁠, and the limits become $t = \sqrt{2}$ and $t = 2$ ⁠.

Given, to find, and the plan

Given: The integrand $x\sqrt{x+2}$ on $[0, 2]$ ⁠, and the hint $x + 2 = t^2$ ⁠.
To find: The exact value of the integral, and a decimal.
Plan: (1) Check the integrand is defined and continuous. (2) Choose $t = \sqrt{x+2} \ge 0$ ⁠; write $x$ ⁠, $dx$ ⁠, $\sqrt{x+2}$ and the new limits. (3) Integrate the polynomial in $t$ ⁠. (4) Evaluate, simplify to the book's form. (5) Check with an antiderivative in $x$ ⁠, and numerically.

Basics you need

1 $\sqrt{t^2} = |t|$

The square root sign means the non-negative root, so $\sqrt{t^2} = t$ only when $t \ge 0$ ⁠. That is why we choose $t \ge 0$ ⁠. Basics: powers and roots

2Substitution with new limits

With $x = g(t)$ running from $a$ to $b$ as $t$ runs from $\alpha$ to $\beta$ ⁠: $\int_a^b f(x)\,dx = \int_\alpha^\beta f(g(t))\,g'(t)\,dt$ (Section 7.9). Lesson: integration by substitution

3Power rule

$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ (standard formula (i), Section 7.2). Basics: the power rule

4The second fundamental theorem

$\int_a^b f = F(b) - F(a)$ when $f$ is continuous on $[a, b]$ and $F' = f$ (Theorem 2, Section 7.8.3). Lesson: definite integrals

Solution

Check the integrand is defined and continuous

For $0 \le x \le 2$ ⁠, $x + 2$ lies between 2 and 4, so it is positive and $\sqrt{x+2}$ is a real number. $\sqrt{x + 2}$ is continuous wherever $x + 2 \ge 0$ ⁠, and $x$ is a polynomial, so $x\sqrt{x+2}$ is continuous on $[0, 2]$ ⁠.

A square root is only real for a non-negative input. The whole interval keeps $x + 2 \ge 2$ ⁠, well away from 0.

A product of continuous functions is continuous (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem
Substitute $x + 2 = t^2$ with $t \ge 0$

Take the non-negative root: $t = \sqrt{x + 2}$ ⁠. Then
$x = t^2 - 2, \quad dx = 2t\,dt, \quad \sqrt{x + 2} = t.$
New limits. When $x = 0$ ⁠, $t = \sqrt{2}$ ⁠. When $x = 2$ ⁠, $t = \sqrt{4} = 2$ ⁠.

Why it is valid on the whole interval. On $\left[\sqrt{2}, 2\right]$ ⁠, $x = t^2 - 2$ has the continuous derivative $2t > 0$ ⁠, so it increases and runs exactly once from 0 to 2. Because $t \ge 0$ ⁠, $\sqrt{t^2} = t$ with no sign trouble.

The hint $x + 2 = t^2$ is chosen so that the root disappears. Writing both $x$ and $dx$ in terms of $t$ is necessary: the integrand has an $x$ outside the root too.

$\frac{d}{dt}(t^2 - 2) = 2t$ ⁠. Basics: substitution
Rewrite the integral as a polynomial in $t$
$\begin{aligned}\int_0^2 x\sqrt{x + 2}\,dx &= \int_{\sqrt{2}}^{2}(t^2 - 2)\cdot t\cdot 2t\,dt \\ &= \int_{\sqrt{2}}^{2}\left(2t^4 - 4t^2\right)dt.\end{aligned}$
The three factors are $x = t^2 - 2$ ⁠, $\sqrt{x+2} = t$ and $dx = 2t\,dt$ ⁠. Multiplying out: $(t^2 - 2)\cdot 2t^2 = 2t^4 - 4t^2$ ⁠.

Distribute: $(a - b)c = ac - bc$ ⁠. Basics: powers of x
Integrate and evaluate
$\int_{\sqrt{2}}^{2}\left(2t^4 - 4t^2\right)dt = \left[\frac{2t^5}{5} - \frac{4t^3}{3}\right]_{\sqrt{2}}^{2}.$
At $t = 2$ ⁠: $t^5 = 32$ ⁠, $t^3 = 8$ ⁠, so the bracket is
$\frac{64}{5} - \frac{32}{3} = \frac{192 - 160}{15} = \frac{32}{15}.$
At $t = \sqrt{2}$ ⁠: $(\sqrt{2})^5 = 4\sqrt{2}$ and $(\sqrt{2})^3 = 2\sqrt{2}$ ⁠, so
$\frac{8\sqrt{2}}{5} - \frac{8\sqrt{2}}{3} = \frac{24\sqrt{2} - 40\sqrt{2}}{15} = -\frac{16\sqrt{2}}{15}.$
Subtract:
$\frac{32}{15} - \left(-\frac{16\sqrt{2}}{15}\right) = \frac{32 + 16\sqrt{2}}{15}.$
Powers of $\sqrt{2}$ ⁠: $(\sqrt{2})^2 = 2$ ⁠, so $(\sqrt{2})^3 = 2\sqrt{2}$ and $(\sqrt{2})^5 = (\sqrt{2})^4\sqrt{2} = 4\sqrt{2}$ ⁠. The lower-limit value is negative, so subtracting it adds.

Standard formula (i), Section 7.2, then Theorem 2. Basics: the power rule
Match the book's form

Since $32 = 16\sqrt{2}\cdot\sqrt{2}$ ⁠, take $16\sqrt{2}$ out:
$\frac{32 + 16\sqrt{2}}{15} = \frac{16\sqrt{2}}{15}\left(\sqrt{2} + 1\right).$
In decimals: $16\sqrt{2} \approx 22.6274170$ ⁠, so the value is about $\frac{32 + 22.6274170}{15}$ ⁠, which is $\frac{54.6274170}{15} \approx 3.6418278$ ⁠.

Both forms are the same number; the book factors out $16\sqrt{2}$ ⁠. Multiplying back, $16\sqrt{2}\cdot\sqrt{2} = 32$ and $16\sqrt{2}\cdot 1 = 16\sqrt{2}$ ⁠.

$\sqrt{2}\cdot\sqrt{2} = 2$ ⁠. Basics: powers and roots
Same result by staying in $x$

Put $t = \sqrt{x+2}$ back into $\frac{2t^5}{5} - \frac{4t^3}{3}$ ⁠:
$F(x) = \frac{2}{5}(x + 2)^{5/2} - \frac{4}{3}(x + 2)^{3/2}.$
Check: differentiate.
$\begin{aligned}F'(x) &= (x + 2)^{3/2} - 2(x + 2)^{1/2} \\ &= (x + 2)^{1/2}\big[(x + 2) - 2\big] \\ &= x\sqrt{x + 2},\end{aligned}$
the integrand. Then $F(2) = \frac{64}{5} - \frac{32}{3}$ and $F(0) = \frac{8\sqrt{2}}{5} - \frac{8\sqrt{2}}{3}$ ⁠, the same numbers as Step 4, so $F(2) - F(0)$ is the same value.

$\frac{2}{5}\cdot\frac{5}{2} = 1$ and $\frac{4}{3}\cdot\frac{3}{2} = 2$ give the two terms; factoring $(x+2)^{1/2}$ leaves $x$ ⁠.

Chain rule with inner $x + 2$ (derivative 1). Basics: the chain rule backwards
The region

$y = x\sqrt{x+2}$ is negative between $-2$ and 0 (outside our interval) and rises from 0 to 4 on $[0, 2]$ ⁠. The shaded area is $\frac{16\sqrt{2}}{15}(\sqrt{2}+1) \approx 3.642$ ⁠.

On $[0, 2]$ the integrand is $\ge 0$ ⁠, so the value is the shaded area. It must be less than the 2 by 4 rectangle (area 8) and, since the curve bends upward, it lies below the straight line from (0, 0) to (2, 4), so the area is less than that triangle's 4. Our 3.642 fits.

For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

Result

$\int_0^2 x\sqrt{x+2}\,dx = \frac{16\sqrt{2}}{15}\left(\sqrt{2} + 1\right) \approx 3.6418278.$ (Textbook answer: $\frac{16\sqrt{2}}{15}(\sqrt{2} + 1)$ ⁠.)

Check it with numbers

A numerical integration of $x\sqrt{x+2}$ over $[0, 2]$ gives 3.6418278. In the book's form, $\frac{16\sqrt{2}}{15} \approx 1.508494467$ (using $\sqrt{2} \approx 1.414213562$ ⁠) and $\sqrt{2} + 1 \approx 2.414213562$ ⁠; their product is about 3.6418278. (Seven-decimal factors are too coarse here: $1.5084945\times 2.4142136$ is about 3.6418279.)

Common mistakes

Changing only the root. $x$ itself must become $t^2 - 2$ ⁠, and $dx$ must become $2t\,dt$ ⁠.
Wrong lower limit. At $x = 0$ ⁠, $t^2 = 2$ ⁠, so $t = \sqrt{2}$ ⁠, not 2 or 0.
Taking $t = -\sqrt{x+2}$ ⁠. Then $\sqrt{x+2} = -t$ ⁠, not $t$ ⁠. Choosing $t \ge 0$ avoids the sign error.
$(\sqrt{2})^5 = 5\sqrt{2}$ ⁠. Wrong: $(\sqrt{2})^5 = 4\sqrt{2}$ ⁠.

What to take away

For $\sqrt{ax + b}$ ⁠, put $ax + b = t^2$ with $t \ge 0$ ⁠: the root becomes $t$ and everything else becomes a polynomial in $t$ ⁠.
Write all three pieces: the root, the leftover $x$ ⁠, and $dx$ ⁠.
Answers with surds can be written several ways; factor to match the book.

Every fact used, and where it comes from

Fact	Used in	Source
Continuity of $\sqrt{x+2}$ for $x \ge -2$ and of products	Step 1	Chapter 5, Section 5.2.1
Substitution $x + 2 = t^2$ with new limits	Steps 2 and 3	Section 7.9; hint in the question (Lesson)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$	Step 4	Standard formula (i), Section 7.2 (Basics)
$\int_a^b f = F(b) - F(a)$	Steps 4 and 6	Theorem 2, Section 7.8.3 (Lesson)
Chain rule	Step 6	Chapter 5, Section 5.3.1

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Exercise 7.9 · Q4: ∫02xx+2 dx\int_0^2 x\sqrt{x+2}\,dx∫02​xx+2​dx

Basics you need

Solution

Check the integrand is defined and continuous

Substitute x+2=t2x + 2 = t^2x+2=t2 with t≥0t \ge 0t≥0

Rewrite the integral as a polynomial in ttt

Integrate and evaluate

Match the book's form

Same result by staying in xxx

The region

Every fact used, and where it comes from

Exercise 7.9 · Q4: $\int_0^2 x\sqrt{x+2}\,dx$

Substitute $x + 2 = t^2$ with $t \ge 0$

Rewrite the integral as a polynomial in $t$

Same result by staying in $x$