Chapter 7 · Integrals

Example 26: $\int_{-1}^{1}5x^4\sqrt{x^5+1}\,dx$

ExamplesDefinite integrals by substitution · Section 7.9

The question

Evaluate $\displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\; dx$ ⁠.

Exam answer

Let $I = \displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\,dx$ ⁠.

Put $x^5 + 1 = t$ ⁠. Then $5x^4\,dx = dt$ ⁠.

When $x = -1$ ⁠, $t = 0$ ⁠; when $x = 1$ ⁠, $t = 2$ ⁠.

\begin{aligned} I &= \int_0^2\sqrt t\,dt = \frac23\Big[t^{3/2}\Big]_0^2 \\ &= \frac23\left(2^{3/2} - 0\right) = \frac23\cdot 2\sqrt2 \end{aligned}

$\therefore\ I = \dfrac{4\sqrt2}{3}$ ⁠.

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

The one idea this problem is built on

The factor $5x^4$ is exactly the derivative of $x^5 + 1$ ⁠, the quantity under the root. So the substitution $t = x^5 + 1$ ⁠, $dt = 5x^4\,dx$ ⁠, turns the whole integrand into $\sqrt t\,dt$ ⁠. Section 7.9 shows two ways to finish: go back to $x$ and use the old limits, or change the limits to the new variable at once (as $x$ runs from $-1$ to 1, $t$ runs from 0 to 2).

Given, to find, and the plan

Given: The integrand $5x^4\sqrt{x^5+1}$ on $[-1, 1]$ ⁠.
To find: The exact value of the definite integral.
Plan: Check the integrand is defined and continuous on $[-1, 1]$ ⁠; substitute $t = x^5 + 1$ ⁠; integrate $\sqrt t$ ⁠; evaluate both ways; check numerically.

Basics you need

1Substitution in a definite integral

If $t = g(x)$ ⁠, where $g$ has a continuous derivative on $[a, b]$ and $f$ is continuous at every value that $g$ takes there, then $\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt.$ The limits are the values of the new variable at the old limits. Lesson: integration by substitution

2Power rule with a fractional power

$\int t^{\frac12}\,dt = \frac{t^{\frac32}}{\frac32} = \frac23 t^{\frac32} + C$ ⁠. Basics: the power rule

3Where $\sqrt{u}$ is defined

$\sqrt u$ needs $u \ge 0$ ⁠, and it is continuous on $[0, \infty)$ ⁠. Basics: substitution

4Second fundamental theorem

$\int_a^b f(x)\,dx = F(b) - F(a)$ for continuous $f$ with antiderivative $F$ (Theorem 2, Section 7.8.3). Lesson: the fundamental theorem

Solution

Check the integrand on $[-1, 1]$

The function $x^5$ is increasing, so for $-1 \le x \le 1$ we have $-1 \le x^5 \le 1$ ⁠, that is
$0 \le x^5 + 1 \le 2.$
So $\sqrt{x^5 + 1}$ is defined on the whole interval, and as a composition of continuous functions it is continuous there. Multiplying by the polynomial $5x^4$ keeps it continuous.

Theorem 2 (and the change of limits) needs the integrand to be defined and continuous on all of $[a, b]$ ⁠. The left end is right at the edge: $x^5 + 1 = 0$ at $x = -1$ ⁠, so the root is 0 there, which is still allowed. For $x < -1$ the root would not exist.

An odd power keeps the order of numbers: if $x_1 < x_2$ then $x_1^5 < x_2^5$ ⁠. Basics: powers of x
Substitute $t = x^5 + 1$

Then $\frac{dt}{dx} = 5x^4$ ⁠, so $dt = 5x^4\,dx$ ⁠. The integrand is $\sqrt{x^5 + 1}\cdot 5x^4\,dx = \sqrt t\,dt$ ⁠:
$\int 5x^4\sqrt{x^5 + 1}\,dx = \int\sqrt t\,dt = \int t^{\frac12}\,dt.$
This is the pattern $\int f(g(x))\,g'(x)\,dx$ with $g(x) = x^5 + 1$ and $f = \sqrt{\;}$ ⁠. The derivative $g'(x) = 5x^4$ is already present, with the right constant, so nothing is left over.

Integration by substitution, Section 7.3.1: the chain rule read backwards. Basics: substitution
Integrate and return to $x$
$\int t^{\frac12}\,dt = \frac23 t^{\frac32} = \frac23\left(x^5 + 1\right)^{\frac32} = F(x).$
Check: $F'(x) = \frac23\cdot\frac32\left(x^5+1\right)^{\frac12}\cdot 5x^4$ ⁠, which is $5x^4\sqrt{x^5+1}$ ⁠.

Dividing by the new exponent $\frac32$ is the same as multiplying by $\frac23$ ⁠. The check uses the chain rule: the outer power gives $\frac32(\ldots)^{\frac12}$ and the inner function gives $5x^4$ ⁠.

Power rule, standard formula (i) with $n = \frac12$ ⁠. Basics: the power rule
Method 1: old limits
$\begin{aligned}I &= \frac23\left[\left(x^5 + 1\right)^{\frac32}\right]_{-1}^{1} \\ &= \frac23\left[(1 + 1)^{\frac32} - \left((-1)^5 + 1\right)^{\frac32}\right] \\ &= \frac23\left[2^{\frac32} - 0^{\frac32}\right] = \frac23\cdot 2\sqrt2 = \frac{4\sqrt2}{3}.\end{aligned}$
$2^{\frac32} = 2\cdot 2^{\frac12} = 2\sqrt2$ ⁠, and $(-1)^5 = -1$ ⁠, so the lower bracket is $0^{\frac32} = 0$ ⁠.

Second fundamental theorem: F(upper) minus F(lower). Lesson: the fundamental theorem
Method 2: change the limits

When $x = -1$ ⁠, $t = (-1)^5 + 1 = 0$ ⁠; when $x = 1$ ⁠, $t = 1 + 1 = 2$ ⁠. So as $x$ runs from $-1$ to 1, $t$ runs from 0 to 2, and
$I = \int_0^2\sqrt t\,dt = \frac23\left[t^{\frac32}\right]_0^2 = \frac23\left(2\sqrt2 - 0\right) = \frac{4\sqrt2}{3}.$
This is the quicker route recommended in the Note of Section 7.9: no need to go back to $x$ ⁠. It is legitimate because $t = x^5 + 1$ has the continuous derivative $5x^4$ on $[-1, 1]$ ⁠, its values fill exactly $[0, 2]$ (Step 1), and $\sqrt t$ is continuous on $[0, 2]$ ⁠.

New limits = values of the new variable at the old limits. Lesson: integration by substitution
The region

$y = 5x^4\sqrt{x^5+1}$ on $[-1, 1]$ ⁠. It is never negative, is 0 at $x = -1$ and $x = 0$ ⁠, and reaches $5\sqrt2 \approx 7.07$ at $x = 1$ ⁠. The shaded area is $\frac{4\sqrt2}{3} \approx 1.886$ ⁠.

The value is $\frac{4\sqrt2}{3} \approx 1.8856181$ ⁠.

The integrand is $\ge 0$ on the whole interval (a fourth power times a root), so the integral is a genuine area. The small hump on the left and the steep rise on the right together make about 1.886.

Definite integral of a non-negative function = area under its graph. Basics: area as a sum of thin strips

Result

$\int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\,dx = \frac{4\sqrt2}{3} \approx 1.8856181,$ the same as the book's answer, by both methods.

Check it with numbers

A numerical integration of $5x^4\sqrt{x^5+1}$ over $[-1, 1]$ gives 1.8856181, and $\frac{4\sqrt2}{3} = \frac{4\times 1.41421356}{3} \approx 1.8856181$ ⁠. Split by pieces: $\int_{-1}^{0} = \frac23(1 - 0) = \frac23 \approx 0.6666667$ and $\int_0^1 = \frac23\left(2\sqrt2 - 1\right) \approx 1.2189514$ ⁠, which add up to 1.8856181.

Common mistakes

Keeping the limits $-1$ and 1 after switching to $t$ ⁠. In $t$ the limits are 0 and 2.
Thinking an even-looking factor makes it symmetric. $5x^4$ is even, but $\sqrt{x^5+1}$ is not, so the even-function property P7 of Section 7.10 does not apply; the two halves are $\frac23$ and about 1.219.
$\int\sqrt t\,dt = \frac32 t^{\frac32}$ ⁠. Divide by the new power: the factor is $\frac23$ ⁠.
$(-1)^5 = 1$ ⁠. An odd power of $-1$ is $-1$ ⁠.

What to take away

Look for "a function and its derivative" in the integrand: here $x^5 + 1$ and $5x^4$ ⁠.
In a definite integral, change the limits along with the variable and you never need to come back to $x$ ⁠.
Check the integrand exists on the whole interval: a square root at the edge of its domain (here at $x = -1$ ⁠) is fine, beyond it is not.

Every fact used, and where it comes from

Fact	Used in	Source
$x^5$ increasing, so $0 \le x^5 + 1 \le 2$	Step 1	Odd powers preserve order
$t = x^5 + 1,\ dt = 5x^4\,dx$	Step 2	Section 7.3.1 (Lesson)
$\int t^{\frac12}dt = \frac23 t^{\frac32} + C$	Step 3	Standard formula (i), Section 7.2 (Basics)
$\int_a^b f = F(b) - F(a)$	Step 4	Theorem 2, Section 7.8.3 (Lesson)
Changing the limits after substituting	Step 5	Section 7.9 and its Note

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Example 26: ∫−115x4x5+1 dx\int_{-1}^{1}5x^4\sqrt{x^5+1}\,dx∫−11​5x4x5+1​dx

Basics you need

Solution

Check the integrand on [−1,1][-1, 1][−1,1]

Substitute t=x5+1t = x^5 + 1t=x5+1

Integrate and return to xxx

Method 1: old limits

Method 2: change the limits

The region

Every fact used, and where it comes from

Example 26: $\int_{-1}^{1}5x^4\sqrt{x^5+1}\,dx$

Check the integrand on $[-1, 1]$

Substitute $t = x^5 + 1$

Integrate and return to $x$