Q. 78 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

${\int }_0^{1}x\sqrt{1-x}dx$

Step-by-Step Solution

Answer

The solution of the given integral is ${\int }_0^{1}x\sqrt{1-x}dx=\frac{2}{3}$.

Step 1. Given Information

Solving the given integrals.

${\int }_0^{1}x\sqrt{1-x}dx$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=1-x \\ \frac{du}{dx}=-1 \\ du=-dx \\ -du=dx \end{gathered}$$

Step 3. We will now write the limits of integration in terms of the new variable u .

When $x=0$, we have

$$\begin{gathered} u=1-x \\ u=1-0 \\ u=1 \end{gathered}$$

When $x=1$, we have

$$\begin{gathered} u=1-x \\ u=1-1 \\ u=0 \end{gathered}$$

Step 4. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_0^{1}x\sqrt{1-x}dx=-{\int }_1^0\sqrt{u}du \\ {\int }_0^{1}x\sqrt{1-x}dx=-{\int }_1^0{u}^{1/2}du \\ {\int }_0^{1}x\sqrt{1-x}dx=-{\left[\frac{u^{1/2+1}}{1/2+1}\right]}_1^0 \\ {\int }_0^{1}x\sqrt{1-x}dx=-{\left[\frac{u^{3/2}}{3/2}\right]}_1^0 \\ {\int }_0^{1}x\sqrt{1-x}dx=-\frac{2}{3}{\left[u^{3/2}\right]}_1^0 \\ {\int }_0^{1}x\sqrt{1-x}dx=-\frac{2}{3}\left[{\left(0\right)}^{3/2}-{\left(1\right)}^{3/2}\right] \\ {\int }_0^{1}x\sqrt{1-x}dx=-\frac{2}{3}\left[0-1\right] \\ {\int }_0^{1}x\sqrt{1-x}dx=-\frac{2}{3}(-1) \\ {\int }_0^{1}x\sqrt{1-x}dx=\frac{2}{3} \end{gathered}$$