# Read This Simple Work for Definite Integrals by U-SUBSTITUTION Method Explained in 3 Distinct Examples and Video Lessons

U-substitution in definite integrals is very similar to the method in indefinite integrals, while the initial bounds (usually x-values) need to be changed to the corresponding u-values for upper and lower limits. You need to account for the limits of the integration.

Alternatively, you can integrate the integral expressions using u-substitutions, then change u-expressions back to x-expressions before substituting the x-bounds.

However it is your choice to perform the definite integrals, we use the method which changing all x-values to u-values in this article.

$$\displaystyle \int_{x_1}^{x_2} f(x) dx \leadsto \int_{u_1}^{u_2} f(u) du$$

## Basic Form of Definite Integrals by u-Substitution

Find $\displaystyle \int_{-1}^{0} x(1+x)^{10} dx$ using $u = 1+x$.

\displaystyle \begin{align} u &= 1+x &\text{Step 1: Letting the substitution} \\ \frac{d}{dx}u &= \frac{d}{dx} (1+x) &\text{Step 2: Derive the substitution} \\ \frac{du}{dx} &= 1 \\ du &= dx \\ u &= 1+x \leadsto x = u-1 &\text{Step 3: Change the subject} \\ x &=0 \leadsto u=1+0 = 1 &\text{Step 4: Change the bounds} \\ x &= -1 \leadsto u = 1-1=0 \\ \int_{-1}^{0} x(1+x)^{10} dx &= \int_{0}^{1} (u-1) u^{10} du \\ &= \int_{0}^{1} (u^{11} – u^{10}) du &\text{Step 5: Replace all } x \text{ to } u \\ &= \left[\frac{1}{12}u^{12} – \frac{1}{11} u^{11} \right]_{0}^{1} &\text{Step 6: Evaluate definite integral} \\ &= \frac{1}{12} – \frac{1}{11} \\ &= -\frac{1}{132} \end{align}

## Definite Integrals involving Surds

Find $\displaystyle \int_{0}^{1} x \sqrt{1-x^2} dx$ using $u = 1 -x^2$.

\displaystyle \begin{align} \frac{d}{dx} u &= \frac{d}{dx} (1-x^2) \\ \frac{du}{dx} &= -2x \\ \frac{du}{-2x} &= dx \\ x &= 1 \leadsto u = 1-1^2 = 0 \\ x &= 0 \leadsto u=1-0^2 = 1 \\ \int_{0}^{1} x \sqrt{1-x^2} dx &= \int_{1}^{0} x \sqrt{u} \frac{du}{-2x} \\ &= -\frac{1}{2} \int_{1}^{0} \sqrt{u} du \\ &= \frac{1}{2} \int_{0}^{1} \sqrt{u} du \\ &= \frac{1}{2} \int_{0}^{1} u^{\frac{1}{2}} du \\ &= \frac{1}{2} \times \frac{1}{\frac{1}{2}+1} \left[u^{\frac{1}{2}+1}\right]_{0}^{1} \\ &= \frac{1}{2} \times \frac{1}{\frac{3}{2}} \left[u^{\frac{3}{2}}\right]_{0}^{1} \\ &= \frac{1}{3} \left[u^{\frac{3}{2}}\right]_{0}^{1} \\ &= \frac{1}{3} \left[ \sqrt{u^3} \right]_{0}^{1} \\ &= \frac{1}{3} \left[ \sqrt{1^3} – \sqrt{0^3} \right] \\ &= \frac{1}{3} \end{align}

The following video explains further with detailed explains.

## Definite Integrals by Squaring Substitutions

Find $\displaystyle \int_{3}^{18} \frac{x}{\sqrt{x-2}} dx$ using $u = \sqrt{x-2}$.

\begin{align} \displaystyle u^2 &= x-2 \\ x &= u^2 + 2 \\ \frac{d}{du} x &= \frac{d}{du} (u^2 + 2) \\ \frac{dx}{du} &= 2u \\ dx &= 2udu \\ x &= 18 \leadsto u = \sqrt{18-2} = 4 \\ x &= 3 \leadsto u = \sqrt{3-2} = 1 \\ \int_{3}^{18} \frac{x}{\sqrt{x-2}} dx &= \int_{1}^{4} \frac{u^2 + 2}{u} 2udu \\ &= \int_{1}^{4} (2u^2 + 4) du \\ &= \left[ \frac{2}{3} u^3 + 4u \right]_{1}^{4} \\ &= \left[ \frac{2}{3} \times 4^3 + 4 \times 4 \right] – \left[ \frac{2}{3} \times 1^3 + 4 \times 1 \right] \\ &= 54 \end{align}

You can watch the following video lesson for consolidating your knowledge.

Now, you can try the following question to test your knowledge.

### Can you use u-substitution for definite integrals?

Absolutely! You can apply the u-substitution method for evaluating definite integrals. While some questions provide the substitutions, you will sometimes need to find the right substitutions by yourself.

### How do you change bounds of definite integrals with u-substitution?

Depending on the u-substitutions, it is required to alter x bounds to corresponding u bounds. For example, $\displaystyle \int_{1}^{2} f(x) dx$ with a substitution $u = x+10$ will change to $\displaystyle \int_{1+10}^{2+10} f(u) du$.

### How do you express definite integrals in terms of u?

The first step is to find the derivative of the u-substitution. For instance, look at an example to find $\displaystyle \int_{1}^{2} x(x^2+1)^{6} dx$ using $u=x^2+1$.
Differentiate both sides of $u=x^2+1$, say $\displaystyle \frac{d}{dx} u = \frac{d}{dx} (x^2+1)$. This gives you $\displaystyle \frac{du}{dx} = 2x$.
Then you can obtain $dx = \displaystyle \frac{du}{2x}$, which means $dx$ is to be replaced by $\displaystyle \frac{du}{2x}$.

Therefore $\displaystyle \int_{1}^{2} x(x^2+1)^{6} dx = \int_{2}^{5} x u^{6} \frac{du}{2x} = \frac{1}{2} \int_{2}^{5} u^{6} du$. 