Q. 33 - Calculus

Question

Definite integrals: Use the Fundamental Theorem of Calculus to find the exact values of each of the definite integrals that follow. Sketch the areas described by these definite integrals to determine whether your answers are reasonable. 

${\int }_{-1}^1\frac{x}{1+x^2}dx$

Step-by-Step Solution

Answer

The solution is: $0$.

Step 1: Given

The given integral is:  ${\int }_{-1}^1\frac{x}{1+x^2}dx$

Step 2: To Find

Use the Fundamental Theorem of Calculus to find the exact values of each of the definite integrals that follow. Sketch the areas described by these definite integrals to determine whether your answers are reasonable.  

Step 3: Calculation

Let,

$$\begin{gathered} u=1+x^2 \\ du=2xdx \\ \frac{du}{2x}=dx \end{gathered}$$

If $x=-1$, then $u=1+{\left(-1\right)}^2=2$

If $x=1$, then $u=1+{\left(1\right)}^2=2$

Using substitution we get:

$$\begin{gathered} {\int }_{-1}^1\frac{x}{1+x^2}dx \\ ={\int }_2^2\frac{x}{u}\frac{du}{2x} \\ =\frac{1}{2}{\int }_2^2\frac{1}{u}du \\ =\frac{1}{2}{\left[\ln \left|u\right|\right]}_2^2 \\ =\frac{1}{2}\left[\ln \left|2\right|-\ln \left|2\right|\right] \\ =\frac{1}{2}\left[0\right] \\ =0 \end{gathered}$$

The sketch of the region is: 

 

Since the shaded regions are equal and opposite to each other, so our answer seems reasonable.