f(x)=√(x + 7) f(x)=f(x) ∫₂⁹ f(x)dx= related content the fundamental theorem of calculus and definite integrals

Explanation:

Step1: Find the derivative of $F(x)$

Using the chain - rule, if $F(x)=\sqrt{x + 7}=(x + 7)^{\frac{1}{2}}$, then $f(x)=F^{\prime}(x)=\frac{1}{2}(x + 7)^{-\frac{1}{2}}\times1=\frac{1}{2\sqrt{x+7}}$.

Step2: Apply the fundamental theorem of calculus

The fundamental theorem of calculus states that if $F^{\prime}(x)=f(x)$, then $\int_{a}^{b}f(x)dx=F(b)-F(a)$. Here, $a = 2$, $b = 9$, and $F(x)=\sqrt{x + 7}$. So $\int_{2}^{9}f(x)dx=F(9)-F(2)$.

Step3: Calculate $F(9)$ and $F(2)$

$F(9)=\sqrt{9 + 7}=\sqrt{16}=4$, $F(2)=\sqrt{2+7}=\sqrt{9}=3$.

Step4: Find the value of the definite integral

$\int_{2}^{9}f(x)dx=F(9)-F(2)=4 - 3=1$.

Answer:

$1$