Substitution Rule for Definite Integrals — Question 4

Question 4

Evaluate the definite integral $$\int_{0}^{1} \frac{x^3}{\sqrt{1+x^4}}\,dx.$$

Question 4 - Solution

This integral suggests a substitution involving the expression inside the square root.

Let $$u=1+x^4.$$ Then $$du=4x^3\,dx \quad\Rightarrow\quad x^3\,dx=\frac14\,du.$$

Change the limits of integration. When $x=0$, $u=1$. When $x=1$, $u=2$.

Substitute into the integral: $$\int_{0}^{1} \frac{x^3}{\sqrt{1+x^4}}\,dx = \frac14\int_{1}^{2} u^{-1/2}\,du.$$

Integrate: $$\frac14\int u^{-1/2}\,du = \frac14\cdot 2u^{1/2} = \frac12\sqrt{u}.$$

Apply the limits: $$\frac12\sqrt{u}\Big|_{1}^{2} = \frac12(\sqrt{2}-1).$$

$$\boxed{\frac{\sqrt{2}-1}{2}}$$