Chapter 5, Problem 40RE

To determine

To calculate: The derivative of the integral value of the integration.

Answer to Problem 40RE

The derivative of the integral value is $\frac{x^2}{1+x^3}$ .

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^3}dx}$ .

Formula used:

The standard Integral of

  ${\displaystyle \int \frac{dx}{x}=\log x+c}$ .

Calculation:

The integration can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^3}dt}\\ {\text{let u = 1+t}}^3\text{ then}\\ {\text{du = 3t}}^2\text{dt then substituting the value,}\\ \begin{array}{cc}\begin{array}{cc}t& 0\end{array}& x\end{array}\\ \begin{array}{ccc}u& 1& {\text{1+x}}^3\end{array}\\ {\displaystyle \underset{1}{\overset{1+x^3}{\int }}\frac{1}{3}\frac{du}{u}}\\ =\frac{1}{3}{\displaystyle \underset{1}{\overset{1+x^3}{\int }}\frac{du}{u}}\\ =\frac{1}{3}{\left[\log u\right]}_1^{1+x^3}\\ =\frac{1}{3}\left[\log \left(1+x^3\right)+\log 1\right]\\ =\frac{1}{3}\left[\log \left(1+x^3\right)+0\right]\\ =\frac{\log \left(1+x^3\right)}{3}\\ \text{the derivative of the integral is,}\\ \frac{dy}{dx}=\frac{1}{3}\cdot \frac{1}{1+x^3}\cdot 3x^2\\ \frac{dy}{dx}=\frac{x^2}{1+x^3}\end{array}$