Chapter 6.4, Problem 24E

To determine

To Construct: Function of the form $y={{\displaystyle \int }}_a^{x}f(t)dt+C$ that satisfies the given condition.

Answer to Problem 24E

Function of the form ${{\displaystyle \int }}_{-3}^x[\sqrt{3-\cos t}]dt+4$

Explanation of Solution

Given information :

Given that $\frac{dy}{dx}=\sqrt{3-\cos x},$ , and $y=4$ when $x=-3$ .

Calculation :

The Fundamental Theorem of Calculus, part 1:

If f is continuous on $[a,b]$ , then the function

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

has a derivative at every point $x$ in $[a,b],$ and

  $\frac{dF}{dx}=\frac{d}{dx}{{\displaystyle \int }}_a^{x}f(t)dt=f(x)$

So from the Fundamental theorem of calculus,

  $f(x)=\frac{dy}{dx}=\sqrt{3-\cos x}$ .............(2)

Now, to construct a function of the form $y={{\displaystyle \int }}_a^{x}f(t)dt+C$

From equation (2), $f(t)=\sqrt{3-\cos t}$

  $y={{\displaystyle \int }}_a^{x}f(t)dt+C$

  $={{\displaystyle \int }}_a^x[\sqrt{3-\cos t}]dt+C\dots \dots$ (3)

Using initial conditions, $y=4$ when $x=-3$ .

Substitute these values in (3),

4= NA+C .........(4)

  ${{\displaystyle \int }}_a^{a}f(x)dx=0$

If $a=-3,{{\displaystyle \int }}_{-3}^{-3}[\sqrt{3-\cos t}]dt=0$ ,

Now substitute this value in (4),

  $4={{\displaystyle \int }}_{-3}^{-3}[\sqrt{3-\cos t}]dt+C$

  $\Rightarrow 4=0+C$

  $\Rightarrow C=4$

Now substitute $\text{C}=4,a=-3$ , in the equation (3),

Hence, ${{\displaystyle \int }}_{-3}^x[\sqrt{3-\cos t}]dt+4$

  $$ .