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F(x)=sqrt(x+7)

f(x)=F^(')(x)

int_(2)^(9)f(x)dx=

$F(x)=\sqrt{x+7}$ $\newline$ $f(x)=F^{\prime}(x)$ $\newline$ $\int_{2}^{9} f(x) d x=$

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Evaluate definite integrals using the power rule

Full solution

Q.

F(x)=\sqrt{x+7}

\newline

f(x)=F^{\prime}(x)

\newline

\int_{2}^{9} f(x) d x=

Identify Function: $F(x) = \sqrt{x+7}$ , so let's find $F'(x)$ which is $f(x)$ .
Find Derivative: Using the chain rule, $F'(x) = \frac{1}{2\sqrt{x+7}} \cdot (x+7)' = \frac{1}{2\sqrt{x+7}} \cdot 1$ .
Calculate Integral: Now we have $f(x) = \frac{1}{2\sqrt{x+7}}$ . Let's integrate $f(x)$ from $2$ to $9$ .
Evaluate Upper Limit: $\int_{2}^{9} \frac{1}{2\sqrt{x+7}} \, dx = [\sqrt{x+7}]$ from $2$ to $9$ .
Evaluate Lower Limit: Plug in the upper limit: $\sqrt{9+7} = \sqrt{16} = 4$ .
Subtract Limits: Plug in the lower limit: $\sqrt{2+7} = \sqrt{9} = 3$ .
Subtract Limits: Plug in the lower limit: $\sqrt{2+7} = \sqrt{9} = 3$ .Now subtract the lower limit from the upper limit: $4 - 3 = 1$ .

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