1. find the area under the graph of f(x) = 1/5 x^2 + 3 over the interval 2,5.

Explanation:

Step1: Recall definite - integral formula

The area $A$ under the curve $y = f(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}f(x)dx$. Here, $a = 2$, $b = 5$ and $f(x)=\frac{1}{5}x^{2}+3$, so $A=\int_{2}^{5}(\frac{1}{5}x^{2}+3)dx$.

Step2: Use integral rules

By the sum - rule of integration $\int_{a}^{b}(u(x)+v(x))dx=\int_{a}^{b}u(x)dx+\int_{a}^{b}v(x)dx$, we have $A=\int_{2}^{5}\frac{1}{5}x^{2}dx+\int_{2}^{5}3dx$. Also, by the constant - multiple rule $\int_{a}^{b}cu(x)dx=c\int_{a}^{b}u(x)dx$, so $A=\frac{1}{5}\int_{2}^{5}x^{2}dx+3\int_{2}^{5}dx$.

Step3: Apply power - rule of integration

The power - rule is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). Then $\frac{1}{5}\int_{2}^{5}x^{2}dx=\frac{1}{5}\left[\frac{x^{3}}{3}\right]{2}^{5}=\frac{1}{15}(x^{3})\big|{2}^{5}=\frac{1}{15}(5^{3}-2^{3})=\frac{1}{15}(125 - 8)=\frac{117}{15}$. And $3\int_{2}^{5}dx=3[x]_{2}^{5}=3(5 - 2)=9$.

Step4: Calculate the area

$A=\frac{117}{15}+9=\frac{117+135}{15}=\frac{252}{15}=\frac{84}{5}=16.8$.

Answer:

$16.8$