Substitution and the properties of definite integrals

What you will be able to do

• change the limits when you substitute in a definite integral;
• use the properties to split an interval and to handle $|x|$ and absolute values;
• use even and odd symmetry, and the reflection $x \to a + b - x$.

Substitution with new limits

For an indefinite integral you return to $x$ at the end. For a definite integral there is a shorter way. When you put $u = g(x)$, change the limits at the same time. The lower limit $a$ becomes $g(a)$, and the upper limit $b$ becomes $g(b)$.

Then you never need to go back to $x$.

Example 1. Find $\displaystyle\int_0^1 x\,e^{x^2}\,dx$.

1. Let $u = x^2$. Then $x\,dx = \tfrac{1}{2}du$. When $x = 0$, $u = 0$. When $x = 1$, $u = 1$.
2. The integral is $\tfrac{1}{2}\displaystyle\int_0^1 e^u\,du = \tfrac{1}{2}(e - 1) = 0.8591$.

Example 2. Find $\displaystyle\int_0^{\pi/2}\sin x\cos^3 x\,dx$.

1. Let $u = \cos x$. Then $du = -\sin x\,dx$. When $x = 0$, $u = 1$. When $x = \dfrac{\pi}{2}$, $u = 0$.
2. The integral is $-\displaystyle\int_1^0 u^3\,du = \displaystyle\int_0^1 u^3\,du = \dfrac{1}{4}$.

Example 3. Find $\displaystyle\int_0^3 \dfrac{x}{\sqrt{x + 1}}\,dx$. Replace $x$ as well.

1. Let $u = x + 1$, so $x = u - 1$. The limits become $u = 1$ and $u = 4$.
2. The integral is $\displaystyle\int_1^4 \left(u^{1/2} - u^{-1/2}\right)du = \left[\dfrac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4 = \dfrac{8}{3}$.

A warning: $g'$ must be continuous on the interval, and $f$ must be continuous on all the values that $g$ takes. Otherwise the new limits may hide a mistake.

The properties

Here $f$ is continuous on the interval.

• P1. $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$. The name of the variable does not matter.
• P2. $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$. Swapping the limits changes the sign. Also $\displaystyle\int_a^a f(x)\,dx = 0$.
• P3. $\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$. This is the way to handle $|x|$, $[x]$ and piecewise functions.
• P4. $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$.
• P5. $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$. This is P4 with the lower limit $0$.
• P6. $\displaystyle\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f$ is even, and $0$ if $f$ is odd.
• P7. $\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f(2a - x) = f(x)$, and $0$ if $f(2a - x) = -f(x)$.

An even function has $f(-x) = f(x)$, and its graph is symmetric about the $y$-axis, as for $x^2$ and $\cos x$. An odd function has $f(-x) = -f(x)$, and its graph is symmetric about the origin, as for $x^3$ and $\sin x$.

P4 comes from the substitution $u = a + b - x$. It reflects the graph in the middle of the interval. The simulation shows an important case.

Worked examples

Example 4 (P3 with a modulus). Find $\displaystyle\int_{-1}^2 |x|\,dx$.

1. $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$. Split at $0$.
2. $\displaystyle\int_{-1}^0 (-x)\,dx + \int_0^2 x\,dx = \dfrac{1}{2} + 2 = \dfrac{5}{2}$.

Example 5 (P6). Find $\displaystyle\int_{-2}^2 \left(x^3 + x^2\right)dx$.

1. $x^3$ is odd, so its integral is $0$. $x^2$ is even.
2. The value is $2\displaystyle\int_0^2 x^2\,dx = 2 \times \dfrac{8}{3} = \dfrac{16}{3}$.

Example 6 (P4). Find $I = \displaystyle\int_0^3 \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{3 - x}}\,dx$. The antiderivative is not simple, but the symmetry helps.

1. Here $a + b = 3$. By P4, $I = \displaystyle\int_0^3 \dfrac{\sqrt{3 - x}}{\sqrt{3 - x} + \sqrt{x}}\,dx$.
2. Add the two forms of $I$. The integrands add up to $1$. So $2I = \displaystyle\int_0^3 1\,dx = 3$, and $I = \dfrac{3}{2}$.

Practise with new numbers

Try it in Python

The program checks three properties with thin strips. First an odd function, then an even function, then the integral of Example 6.

Common mistakes

• Not changing the limits. After $u = g(x)$, the limits are values of $u$, not of $x$.
• Using even-odd symmetry on a non-symmetric interval. P6 needs the interval $[-a, a]$.
• Missing a sign change. For $|f(x)|$, split where $f$ changes sign.
• Forgetting that P4 needs $a + b$. The reflected integrand is $f(a + b - x)$, not $f(a - x)$ unless $a = 0$.
• Calling a function odd or even without checking. Test $f(-x)$ for each term.

Remember these

• Substitution: change the limits with the variable.
• $\int_a^b f = \int_a^c f + \int_c^b f$ lets you split at a sign change or a joining point.
• Even: $2\int_0^a f$. Odd: $0$.
• P4: $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$, and adding the two forms often gives the answer.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\int_{-\pi/2}^{\pi/2} x^2\sin x\,dx$.

2. Find $\displaystyle\int_0^4 \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{4 - x}}\,dx$.

3. Find $\displaystyle\int_{-1}^3 |x - 1|\,dx$.

4. Find $\displaystyle\int_0^\pi |\cos x|\,dx$.

5. Find $\displaystyle\int_0^2 x\,(x^2 + 1)^3\,dx$ by changing the limits.

References

• OpenStax, Calculus Volume 1, Chapter 5, sections 5.2 and 5.5 (substitution for definite integrals).
• NCERT, Mathematics Part II for Class XII, Chapter 7, the sections on evaluation of definite integrals by substitution and some properties of definite integrals.