evaluate the integral.
int_{1}^{5}(2 x+8) d x
int_{1}^{5}(2 x+8) d x=square

Question

evaluate the integral.
 int_{1}^{5}(2 x+8) d x
 int_{1}^{5}(2 x+8) d x=square

Accepted Answer

# Explanation:
## Step1: Find the antiderivative
The antiderivative of $2x$ is $x^{2}$ (using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$, here $n = 1$), and the antiderivative of $8$ is $8x$. So the antiderivative of $2x + 8$ is $F(x)=x^{2}+8x$.
## Step2: Apply the fundamental theorem of calculus
By the fundamental theorem of calculus $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where $a = 1$, $b = 5$, $F(x)=x^{2}+8x$.
$$
\begin{align*}
F(5)-F(1)&=(5^{2}+8\times5)-(1^{2}+8\times1)\
&=(25 + 40)-(1+8)\
&=65 - 9
\end{align*}
$$

# Answer:
$56$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the antiderivative

Step2: Apply the fundamental theorem of calculus

Answer: