Question

Find the particular antiderivative F(x) of f(x)=8/sqreroot xthat satisfies F(1) = 9.

Answer 1

Given:

$f(x)=\frac{8}{\sqrt[]{x}}$

The above can be re-written as;

$f(x)=\frac{8}{x^{\frac{1}{2}}}=8x^{-\frac{1}{2}}$$f(x)=8x^{-\frac{1}{2}}$

Applying the rule of antiderivatives of basic function;

$x^ndx=\frac{x^{n+1}}{n+1}$

Then we have;

$F(x)=\frac{8x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$F(x)=\frac{8x^{\frac{1}{2}}}{\frac{1}{2}}$$=8x^{\frac{1}{2}}\times2$$=16x^{\frac{1}{2}}+C$

Hence;

$F(x)=16x^{\frac{1}{2}}+C$

But from the question; F(1) = 9

Substitute x =1 into F(x) and equate to 9 to find C.

That is;

$F(1)=16(1)^{\frac{1}{2}}+C=9$

⇒ 16 + C = 9

Subtract 16 from both-side of the equation.

C = 9 - 16

C = -7

Substitute the value of C back into the anti derivative F(x).

Therefore,

The antiderivative that satisfies F(x) of the function that satisfies F(1) = 9 is:

$F(x)=16x^{\frac{1}{2}}-7$