Use the Fundamental Theorem of Calculus to find the exact value of [tex]\int_2^5 \frac{3}{\sqrt{x}} \, dx[/tex].

A) [tex]3(\sqrt{5}-\sqrt{2})[/tex]
B) [tex]2(\sqrt{5}-\sqrt{2})[/tex]
C) [tex]6(\sqrt{5}-\sqrt{2})[/tex]
D) [tex]2(\sqrt{2}-\sqrt{5})[/tex]
E) [tex]6(\sqrt{2}-\sqrt{5})[/tex]
F) None of the above



Answer :

To find the exact value of the definite integral [tex]\(\int_2^5 \frac{3}{\sqrt{x}} \, dx\)[/tex], we will use the Fundamental Theorem of Calculus. Here's a step-by-step solution:

1. Identify the integrand:

The integrand is [tex]\(\frac{3}{\sqrt{x}}\)[/tex].

2. Find the antiderivative:

To find the antiderivative, we need to rewrite the integrand in a more convenient form. Note that [tex]\(\frac{3}{\sqrt{x}} = 3x^{-1/2}\)[/tex].

We use the general power rule for integration: [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex], where [tex]\(n \neq -1\)[/tex].

For [tex]\(n = -\frac{1}{2}\)[/tex]:

[tex]\[ \int 3x^{-1/2} \, dx = 3 \int x^{-1/2} \, dx \][/tex]

[tex]\[ = 3 \cdot \left( \frac{x^{(-1/2)+1}}{(-1/2)+1} \right) + C \][/tex]

[tex]\[ = 3 \cdot \left( \frac{x^{1/2}}{1/2} \right) + C \][/tex]

[tex]\[ = 3 \cdot 2x^{1/2} + C \][/tex]

[tex]\[ = 6\sqrt{x} + C \][/tex]

Therefore, the antiderivative of [tex]\(\frac{3}{\sqrt{x}}\)[/tex] is [tex]\(6\sqrt{x}\)[/tex].

3. Evaluate the definite integral:

We now evaluate the antiderivative at the upper and lower limits and subtract:

[tex]\[ \int_2^5 \frac{3}{\sqrt{x}} \, dx = \left[ 6\sqrt{x} \right]_2^5 \][/tex]

[tex]\[ = 6\sqrt{5} - 6\sqrt{2} \][/tex]

[tex]\[ = 6(\sqrt{5} - \sqrt{2}) \][/tex]

4. Compare with the provided options:

The correct expression is [tex]\(6(\sqrt{5} - \sqrt{2})\)[/tex], which corresponds to option C.

Therefore, the exact value of [tex]\(\int_2^5 \frac{3}{\sqrt{x}} \, dx\)[/tex] is:

[tex]\[ \boxed{6(\sqrt{5} - \sqrt{2})} \][/tex]