Chapter 7, Problem 35RE

To determine

To find the integral equation.

Answer to Problem 35RE

The required solution is, $y={\displaystyle {\int }_a^x{\sin }^{3}t}dt+5$

Explanation of Solution

Given information:The equation is, $\frac{dy}{dx}={\sin }^{3}x$ and $y=5\text{ when }x=4$

Calculation: The required solution is obtained as,

By the fundamental theorem of calculus part 2, ${\displaystyle {\int }_a^{x}f\left(t\right)}dt=F\left(x\right)-F\left(a\right)$

Where F is the antiderivative of f.so,

  $\begin{array}{l}y={\displaystyle {\int }_{}^{}f\left(t\right)}dt+b\\ y-b={\displaystyle {\int }_{}^{}f\left(t\right)}dt+b\end{array}$

Then,

  $y=F\left(x\right)\text{ and }b=F\left(a\right).$

It is given that

  $y=5$ when $x=4$

So,

  $\begin{array}{l}a=4\text{ gives }b=F\left(a\right)\\ b=F\left(4\right)=5\end{array}$

By the fundamental theorem of calculus part 1, $\frac{dy}{dx}=\frac{d}{dx}{\displaystyle {\int }_a^{x}f\left(t\right)}dt=f\left(x\right)$

So,

  $\begin{array}{l}\frac{dy}{dx}={\sin }^{3}x\\ f\left(x\right)={\sin }^{3}x\\ f\left(t\right)={\sin }^{3}t\\ y={\displaystyle {\int }_a^{x}f\left(t\right)}dt+b\\ ={\displaystyle {\int }_a^x{\sin }^{3}t}dt+5\end{array}$

Hence the required solution is, $y={\displaystyle {\int }_a^x{\sin }^{3}t}dt+5$