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 \( u = 1+x \).

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

Please watch the following video for your detailed further study.

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.

Frequently Asked Questions

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

 



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