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]
