Short and Sweet Calculus

## 6.5 Evaluation of Definite Integrals by Substitution

In the previous chapter, we learned the subsitution rule for indefinite integrals. There are two methods to use substitution to evaluate definite integrals.

• The first method is to first find the indefinite integral using substitution and then evaluate the definite integral by appling the second part of the Fundamental Theorem of Calculus.

• The second method is to directly use a substitution in the definite integral by changing both the variable of and the limits of integration in one step, as stated in the following theorem.

6.3. (Substitution in Definite Integrals): If the function $$u=g(x)$$ has a continuous derivative on the interval $$[a,b]$$ and $$f$$ is continuous on the range of $$g$$, then $\int_{a}^{b}f(g(x))g'(x)dx=\int_{u=g(a)}^{u=g(b)}f(u)du.\qquad\left(u=g(x)\right)$

Example 6.10. Evaluate $\int_{0}^{3}x\sqrt{1+x}\ dx.$

Solution

Method (a): An appropriate substitution is $$u=1+x$$ or $$x=u-1$$. Then $dx=du.$ When $$x=0$$, $$u=1+0=1$$ and when $$x=3$$, $$u=1+3=4$$ . Therefore \begin{aligned} \int_{0}^{3}x\sqrt{1+x}\ dx & =\int_{1}^{4}(u-1)\sqrt{u}\ du\\ & =\left[\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right]_{u=1}^{u=4}\\ & =\left(\frac{2}{5}(\sqrt{4})^{5}-\frac{2}{3}(\sqrt{4})^{3}\right)-\left(\frac{2}{5}(1)-\frac{2}{3}(1)\right)\\ & =\frac{116}{15}.\end{aligned} Method (b): Using the substitution $$u=1+x$$, we have \begin{aligned} \int x\sqrt{1+x}\ dx & =\int(u-1)\sqrt{u}\ du\\ & =\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\\ & =\frac{2}{5}(1+x)^{5/2}-\frac{2}{3}(1+x)^{3/2}.\end{aligned} Therefore, \begin{aligned} \int_{0}^{3}x\sqrt{1+x}\ dx & =\left[\frac{2}{5}(1+x)^{5/2}-\frac{2}{3}(1+x)^{3/2}\right]_{x=0}^{x=3}\\ & =\frac{2}{5}(4^{5/2})-\frac{2}{3}(4^{3/2})-\frac{2}{5}+\frac{2}{3}\\ & =\frac{116}{15}.\end{aligned}

• The previous example has been solved by two different methods. In the first method, the transformed definite integral with transformed limits is evaluated using the the Substitution in Definite Integrals (Theorem 6.3). In the second method, first the indefinite integral is evaluated using the Substitution Rule or $$u$$-substituion (Section [sec:Ch7-Integration-by-substitution]) and then original limits are used. There is no general rule to say which method is better and it depends on the problem.