Chapter 6.4, Problem 19E

To determine

To find: The value of $\frac{dy}{dx}$ for the function $y={\displaystyle {\int }_{x^2}^{x^3}\cos 2t}dt$ .

Answer to Problem 19E

The value of $\frac{dy}{dx}=3x^2\cos \left(2x^3\right)-2x\cos \left(2x^2\right)$ .

Explanation of Solution

Given:

The function is $y={\displaystyle {\int }_{x^2}^{x^3}\cos 2t}dt$ .

Concept used:

The fundamental theorem of calculus part 1 can be stated as the derivative of the function $y={\displaystyle {\int }_u^{v}f\left(t\right)dt}$ is $v^{\prime }f\left(v\right)-u^{\prime }f\left(u\right)$ where $u\text{ and }v$ are the function of $x$ .

The power rule of differentiation is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ where $x$ denotes the variable and $n$ is the constant.

Calculation:

The given function is $y={\displaystyle {\int }_{x^2}^{x^3}\cos 2t}dt$ where $u=x^2\text{and }v=x^3$ and $f\left(t\right)=\cos 2t$ .

Differentiate the function $v=x^3$ with respect to $x$ by the use of power rule of differentiation.

  $\begin{array}{c}\frac{dv}{dx}=\frac{d}{dx}\left(x^3\right)\\ =3x^2\end{array}$

Differentiate the function $u=x^2$ with respect to $x$ by the use of power rule of differentiation.

  $\begin{array}{c}\frac{du}{dx}=\frac{d}{dx}\left(x^2\right)\\ =2x\end{array}$

Now find the value of $f\left(u\right)=f\left(x^2\right)$ .

  $\begin{array}{c}f\left(u\right)=f\left(x^2\right)\\ =\cos \left(2x^2\right)\end{array}$

Now find the value of $f\left(v\right)=f\left(x^3\right)$ .

  $\begin{array}{c}f\left(u\right)=f\left(x^3\right)\\ =\cos \left(2x^3\right)\end{array}$

Find the derivative $\frac{dy}{dx}$ by the use of fundamental theorem of calculus.

  $\begin{array}{c}\frac{d}{dx}\left({\displaystyle {\int }_u^{v}f\left(t\right)dt}\right)=v^{\prime }f\left(v\right)-u^{\prime }f\left(u\right)\\ \left({\displaystyle {\int }_{x^2}^{x^3}\cos 2t}dt\right)=3x^2\cos \left(2x^3\right)-2x\cos \left(2x^2\right)\end{array}$

Therefore, the value of $\frac{dy}{dx}=3x^2\cos \left(2x^3\right)-2x\cos \left(2x^2\right)$ .

Conclusion:

Thus, the value of $\frac{dy}{dx}=3x^2\cos \left(2x^3\right)-2x\cos \left(2x^2\right)$ .