Chapter 6.4, Problem 21E

To determine

To find: The function in the form $y={\displaystyle {\int }_a^{x}f\left(t\right)dt+C}$ if $\frac{dy}{dx}={\sin }^{3}x$ and $y=0$ when $x=5$ .

Answer to Problem 21E

The function in the form $y={\displaystyle {\int }_a^{x}f\left(t\right)dt+C}$ is $y={\displaystyle {\int }_5^x{\sin }^{3}dt}$ .

Explanation of Solution

Given:

The derivative of the function is $\frac{dy}{dx}={\sin }^{3}x$ and $y=0$ when $x=5$ .

Concept used:

The fundamental theorem of calculus part 1 can be stated as the derivative of the continuous function at every point $x$ in the interval $\left[a,b\right]$ is $\frac{dy}{dx}=f\left(x\right)$ .

The fundamental theorem of calculus part 2 can be stated as if $f$ is the continuous function on the interval $\left[a,b\right]$ and F is an anti-derivative of $f$ on the interval $\left[a,b\right]$ then it satisfies the condition ${\displaystyle {\int }_a^{b}f\left(t\right)dt}=F\left(b\right)-F\left(a\right)$ .

Calculation:

It is given that $\frac{dy}{dx}={\sin }^{3}x$ .

Use the fundamental theorem of calculus part 1 to obtain the value of $f\left(x\right)$

  $f\left(x\right)={\sin }^{3}x$

Therefore, the value of $f\left(t\right)={\sin }^{3}t$ .

Consider $a=5\text{ and }b=x$ and use the fundamental theorem of calculus part 2 to construct the function.

Substitute $f\left(t\right)={\sin }^{3}t$ and $a=5\text{ and }b=x$ in the fundamental theorem of calculus part 2 to obtain the required function.

  $\begin{array}{c}{\displaystyle {\int }_5^x{\sin }^3\left(t\right)dt}=F\left(x\right)-F\left(5\right)\\ F\left(x\right)={\displaystyle {\int }_5^x{\sin }^3\left(t\right)dt}+F\left(5\right)\end{array}$

It is given that the value of $y=0$ when $x=5$ that means $F\left(5\right)=0$ .

Now solve the function $F\left(x\right)$ as follows:

  $\begin{array}{c}F\left(x\right)={\displaystyle {\int }_5^x{\sin }^3\left(t\right)dt}+0\\ ={\displaystyle {\int }_5^x{\sin }^3\left(t\right)dt}\end{array}$

Therefore, the required function is $F\left(x\right)={\displaystyle {\int }_5^x{\sin }^3\left(t\right)dt}$ .

Conclusion:

Thus, the required function is $y={\displaystyle {\int }_5^x{\sin }^{3}dt}$ .