Chapter 7 · Integrals

Exercise 7.10 · Q17: $\int_0^a\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}\,dx$

Exercise 7.10Properties of definite integrals (P4) · Section 7.10

The question

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

17. $\displaystyle\int_0^{a}\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}\,dx$

Exam answer

\text{Let } I = \int_0^{a}\frac{\sqrt{x}\,dx}{\sqrt{x} + \sqrt{a - x}}\;\ldots(\text{i})

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠, and $a - (a - x) = x$ ⁠,

I = \int_0^{a}\frac{\sqrt{a - x}\,dx}{\sqrt{a - x} + \sqrt{x}}\;\ldots(\text{ii})

Adding (i) and (ii),

\begin{aligned} 2I &= \int_0^{a}\frac{\sqrt{x} + \sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}\,dx \\ &= \int_0^{a}1\,dx = \Big[x\Big]_0^{a} = a \end{aligned}

$\therefore\ I = \dfrac{a}{2}$ (for $a > 0$ ⁠).

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

The one idea this problem is built on

Replacing $x$ by $a - x$ (P4) swaps $\sqrt{x}$ and $\sqrt{a - x}$ ⁠. The new integrand has the same denominator and the "other" square root on top, so the old and new integrands add up to exactly 1. Adding the two forms of $I$ gives $2I = \int_0^a 1\,dx = a$ ⁠.

Given, to find, and the plan

Given: $I = \displaystyle\int_0^{a} f(x)\,dx$ with $f(x) = \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}$ ⁠, where $a > 0$ (needed for the interval $[0, a]$ and for $\sqrt{a - x}$ to make sense).
To find: $I$ in terms of $a$ ⁠.
Plan: (1) Continuity on $[0, a]$ ⁠. (2) P4. (3) Add the two forms. (4) Solve for $I$ ⁠.

Basics you need

1Property P4 (Section 7.10)

$\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ for $f$ continuous on $[0, a]$ ⁠. Lesson: properties of definite integrals

2Square roots

$\sqrt{u}$ is defined and continuous for $u \ge 0$ ⁠, and $\sqrt{u} = 0$ only when $u = 0$ ⁠. Basics: powers of x

3Point symmetry

$f(a - x) = 1 - f(x)$ means the graph is unchanged by a half turn about $(\frac{a}{2}, \frac12)$ ⁠. Basics: symmetry and mirror images

Solution

The integrand is continuous on $[0, a]$

For $0 \le x \le a$ ⁠, both $x \ge 0$ and $a - x \ge 0$ ⁠, so $\sqrt{x}$ and $\sqrt{a - x}$ are defined and continuous. The denominator $\sqrt{x} + \sqrt{a - x}$ is a sum of two numbers $\ge 0$ ⁠; it could be 0 only if both were 0, that is $x = 0$ and $x = a$ at once, which is impossible since $a > 0$ ⁠. So the denominator is positive and $f$ is continuous on $[0, a]$ ⁠.

P4 is stated for continuous functions on $[0, a]$ ⁠. At the ends, $f(0) = 0$ and $f(a) = 1$ ⁠: both are fine, because only one of the two roots vanishes at a time.

Quotients of continuous functions are continuous where the denominator is not 0 (Chapter 5, Section 5.2.1).
P4: replace $x$ by $a - x$

Under $x \to a - x$ ⁠, the expression $a - x$ becomes $a - (a - x) = x$ ⁠. So the two roots swap:
$f(a - x) = \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} ,$
and by P4
$I = \int_0^{a}\frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}\,dx .$
P4 holds for every continuous $f$ (Step 1). We use it because it keeps the denominator the same (addition does not care about order) while putting the other root on top.

P4, Section 7.10. Lesson: properties
Add the two forms

Both integrals run over $[0, a]$ and have the same denominator:
$\begin{aligned}2I &= \int_0^{a}\frac{\sqrt{x} + \sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}\,dx \\ &= \int_0^{a} 1\,dx = a .\end{aligned}$
The numerator equals the denominator, and the denominator is never 0 (Step 1), so the fraction is exactly 1 at every point. The integral of 1 over an interval is its length, here $a - 0 = a$ ⁠.

$\int_a^b 1\,dx = b - a$ ⁠, the area of a rectangle of height 1. Basics: area as a sum of thin strips
Solve for $I$
$I = \frac{a}{2} .$

The case $a = 4$ ⁠. The curve $y = f(x)$ (blue) and $y = f(4 - x) = 1 - f(x)$ (dashed violet) are swapped by a half turn about $(2, \frac12)$ ⁠. The region under the blue curve and the region above it inside the $4 \times 1$ rectangle are congruent, so each is half the rectangle: $I = 2 = \frac{a}{2}$ ⁠.

The figure gives the same answer geometrically: $f(a - x) = 1 - f(x)$ says the region above the curve (inside the $a \times 1$ rectangle) is the region below it turned half a turn, so each has half the area $a \times 1$ ⁠.

Basics: symmetry and mirror images

Result

$\int_0^{a}\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}\,dx = \frac{a}{2} \quad (a > 0)$ (Textbook answer: $\frac{a}{2}$ ⁠.)

Check it with numbers

Numerical integration gives $0.500000000000$ for $a = 1$ ⁠, $2.000000000000$ for $a = 4$ ⁠, and $4.750000000000$ for $a = 9.5$ ⁠, each equal to $\frac{a}{2}$ ⁠.

The pairing of Step 3 for $a = 4$ at $x = 1$ ⁠: $f(1) = \frac{1}{1 + \sqrt3} \approx 0.366025$ and $f(3) = \frac{\sqrt3}{\sqrt3 + 1} \approx 0.633975$ ⁠; they add to 1.

Common mistakes

Forgetting that $a - x$ becomes $x$ under P4, so that only the numerator is changed. Both roots swap; the denominator stays the same only because addition is commutative.
Answering $a$ : the sum of the two forms is $2I$ ⁠, not $I$ ⁠.
Not stating $a > 0$ ⁠. For $a \le 0$ the interval and the roots do not make sense.

What to take away

When P4 turns $f$ into a function $g$ with $f + g$ constant, add the two forms: $I = \frac{1}{2}\int_0^a (f + g)\,dx$ ⁠.
The same pattern, with $\sin x$ and $\cos x$ in place of $\sqrt{x}$ and $\sqrt{a - x}$ ⁠, solves Q2, Q3 and Q4, where the answer is half of $\frac{\pi}{2}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Continuity of roots and of quotients	Step 1	Chapter 5, Section 5.2.1
P4: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$	Step 2	Section 7.10, property P4 (Lesson: properties)
Integral of a sum; $\int_0^a 1\,dx = a$	Step 3	Section 7.2.1, Property (III); Theorem 2, Section 7.8.3

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Exercise 7.10 · Q17: ∫0axx+a−x dx\int_0^a\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}\,dx∫0a​x​+a−x​x​​dx

Basics you need

Solution

The integrand is continuous on [0,a][0, a][0,a]

P4: replace xxx by a−xa - xa−x

Add the two forms

Solve for III

Every fact used, and where it comes from

Exercise 7.10 · Q17: $\int_0^a\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}\,dx$

The integrand is continuous on $[0, a]$

P4: replace $x$ by $a - x$

Solve for $I$