Question

1. \\(\int_{a}^{2a} x \\, dx, \quad a > 0\\) gle sum in sigma in exercises 29–32, express the desired c

Explanation:

Step1: Recall the power rule for integration

The power rule for integration states that $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ (for $n
eq - 1$). In the integral $\int_{a}^{2a}x dx$, we have $n = 1$.

Step2: Apply the fundamental theorem of calculus

The fundamental theorem of calculus part 2 says that if $F(x)$ is an antiderivative of $f(x)$ on $[b,c]$, then $\int_{b}^{c}f(x)dx=F(c)-F(b)$.

First, find the antiderivative of $f(x)=x$. Using the power rule with $n = 1$, the antiderivative $F(x)=\frac{x^{2}}{2}$.

Now, evaluate $F(2a)-F(a)$:

$F(2a)=\frac{(2a)^{2}}{2}=\frac{4a^{2}}{2} = 2a^{2}$

$F(a)=\frac{a^{2}}{2}$

Then $\int_{a}^{2a}x dx=F(2a)-F(a)=2a^{2}-\frac{a^{2}}{2}$

Step3: Simplify the result

$2a^{2}-\frac{a^{2}}{2}=\frac{4a^{2}-a^{2}}{2}=\frac{3a^{2}}{2}$

Answer:

$\frac{3a^{2}}{2}$