Chapter 6, Problem 35RE

To determine

To find: The integral.

Answer to Problem 35RE

The answer is: $y={{\displaystyle \int }}_4^x{\sin }^{3}tdt+5$

Explanation of Solution

Given information:

The equation is: $y={{\displaystyle \int }}_a^{x}f(t)dt+t$

The differential equation is: $dy/dx={\sin }^{3}x$

Calculation:

Part 1 of the fundamental theorem of calculus:

If $f$ is continuous on $(a,b)$ and

  $F(x)={{\displaystyle \int }}_a^{x}f(t)dt$

Then the derivative of $F(x)$ is:

  $\begin{array}{c}{F}^{\prime }(x)=\frac{d}{dx}\left[{{\displaystyle \int }}_a^{x}f(t)dt\right]\\ =f(x)\end{array}$

Hence, $\frac{dy}{dx}={\sin }^3$ use the above theorem to write,

  $y={{\displaystyle \int }}_a^x{\sin }^{3}tdt+C\mathrm{.....}(\text{ Equation }1)$

Put,

  $\begin{array}{c}x=4,\\ y=5\end{array}$

To find $a\text{ ,}C$ :

  $5={{\displaystyle \int }}_a^4{\sin }^{3}tdt+C$

A definite integral is 0 when its upper and lower limits are equal, so set $a=4$ :

  $5=0+C$

In Equation $\text{(1)}$ put back

  $\begin{array}{c}a=4\\ C=5\end{array}$

And get:

  $y={{\displaystyle \int }}_4^x{\sin }^{3}tdt+5$

Therefore the required initial value of the given problem is $y={{\displaystyle \int }}_4^x{\sin }^{3}tdt+5$