Chapter 5.4, Problem 21E

To determine

To find: The function of the form $y={{\displaystyle \int }}_a^{1}f(t)dt+C$ that satisfies the given condition.

Answer to Problem 21E

The answer is: $y={{\displaystyle \int }}_5^x{\sin }^{3}tdt$

Explanation of Solution

Given information:

The condition is:

  $\frac{dy}{dx}={\sin }^{3}x,$

And, $y=0$

When $x=5$

Calculation:

The given condition is:

  $\frac{dy}{dx}={\sin }^{3}x,$

And, $y=0$

When $x=5$

The calculus fundamental theorem, part 1:

If $f$ is continuous on $[a,b]$ and

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

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

  $y={{\displaystyle \int }}_a^x{\sin }^{3}tdt+C$

Put,

  $\begin{array}{c}x=5\\ y=0\end{array}$

To find $a,C$

:

Know that a definite integral is 0 when its upper and lower limits are equal, so set $a=5$

  $0=0+C$

Put back,

  $\begin{array}{c}a=5\\ C=0\end{array}$

To get,

  $y={{\displaystyle \int }}_5^x{\sin }^{3}tdt$

Therefore, the required function of the form $y={{\displaystyle \int }}_a^{1}f(t)dt+C$ that satisfies the given condition is $y={{\displaystyle \int }}_5^x{\sin }^{3}tdt$.