More Questions on Properties of Definite Integrals (Exercise 7.10)

This lesson works through more questions on the properties of definite integrals. The key tools are the substitution that replaces $x$ with $a-x$ over a limit $0$ to $a$, and the odd and even function rules over a symmetric limit.

The complementary substitution

For any integral over $0$ to $a$,

$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx.$

When the integrand is a fraction whose denominator is a symmetric sum, replacing $x$ with $a-x$ and adding the two forms makes the fraction collapse to $1$.

Question 3: sine over sine plus cosine

Let

$I = \int_0^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}\,dx. \qquad (1)$

Replacing $x$ with $\tfrac{\pi}{2}-x$ swaps sine and cosine:

$I = \int_0^{\pi/2} \frac{\cos^{3/2}x}{\cos^{3/2}x + \sin^{3/2}x}\,dx. \qquad (2)$

Adding $(1)$ and $(2)$, the numerators sum to the common denominator:

$2I = \int_0^{\pi/2} 1\,dx = \Big[x\Big]_0^{\pi/2} = \frac{\pi}{2}.$

$I = \frac{\pi}{4}.$

Question 7: x times one minus x to the n

Let

$I = \int_0^1 x\,(1-x)^n\,dx. \qquad (1)$

Replacing $x$ with $1-x$ gives

$I = \int_0^1 (1-x)\,x^n\,dx. \qquad (2)$

Here adding the two forms does not simplify, so we expand $(2)$ and integrate term by term:

$I = \int_0^1 \big(x^n - x^{n+1}\big)\,dx = \left[\frac{x^{n+1}}{n+1}\right]_0^1 - \left[\frac{x^{n+2}}{n+2}\right]_0^1.$

$I = \frac{1}{n+1} - \frac{1}{n+2} = \frac{(n+2)-(n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}.$

Question 9: x times root of two minus x

Let

$I = \int_0^2 x\,\sqrt{2-x}\,dx. \qquad (1)$

Replacing $x$ with $2-x$ gives

$I = \int_0^2 (2-x)\,\sqrt{x}\,dx.$

Expand, writing $\sqrt{x}=x^{1/2}$ and $x\sqrt{x}=x^{3/2}$:

$I = \int_0^2 \big(2x^{1/2} - x^{3/2}\big)\,dx = 2\left[\frac{x^{3/2}}{3/2}\right]_0^2 - \left[\frac{x^{5/2}}{5/2}\right]_0^2.$

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

$I = \frac{4}{3}\,(2\sqrt2) - \frac{2}{5}\,(4\sqrt2) = \frac{8\sqrt2}{3} - \frac{8\sqrt2}{5} = 8\sqrt2\left(\frac{1}{3}-\frac{1}{5}\right).$

$I = 8\sqrt2 \cdot \frac{2}{15} = \frac{16\sqrt2}{15}.$

Question 17: root x over root x plus root a minus x

Let

$I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\,dx. \qquad (1)$

Replacing $x$ with $a-x$,

$I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}}\,dx. \qquad (2)$

Adding $(1)$ and $(2)$, the numerators sum to the common denominator:

$2I = \int_0^a 1\,dx = \Big[x\Big]_0^a = a.$

$I = \frac{a}{2}.$

Question 13: an odd function over a symmetric limit

For a symmetric limit, $\int_{-a}^{a} f(x)\,dx = 0$ when $f$ is odd. With $f(x)=\sin^7 x$,

$f(-x) = \sin^7(-x) = (-\sin x)^7 = -\sin^7 x = -f(x),$

so $f$ is odd and

$\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx = 0.$

Question 20: an odd part plus a constant

Evaluate

$I = \int_{-\pi/2}^{\pi/2} \big(x^3 + x\cos x + \tan^5 x + 1\big)\,dx.$

Split off the constant:

$I = \int_{-\pi/2}^{\pi/2} \big(x^3 + x\cos x + \tan^5 x\big)\,dx + \int_{-\pi/2}^{\pi/2} 1\,dx.$

The first integrand is odd, since $\cos x$ is even (so $x\cos x$ is odd) and $x^3$ and $\tan^5 x$ are odd, so that integral is $0$. The constant part gives

$\int_{-\pi/2}^{\pi/2} 1\,dx = \Big[x\Big]_{-\pi/2}^{\pi/2} = \frac{\pi}{2} + \frac{\pi}{2} = \pi.$

So $I = \pi$.

Key takeaways

• Over $0$ to $a$, replacing $x$ with $a-x$ leaves the integral unchanged; adding the two forms can collapse a symmetric fraction to a constant.
• When adding the two forms does not simplify, use the substituted form to expand and integrate term by term.
• Over a symmetric limit, the integral of an odd function is $0$; separate any constant term and integrate it directly.