Given f(x) = (1)/(x2) + 2x + 3, find the area under the graph of f over the interval [2, 5].

The given function is $f(x) = \frac{1}{5}x^2 + 3$.
The interval is $[2, 5]$, so the lower limit is $a=2$ and the upper limit is $b=5$.

The area under the graph of $f(x)$ from $a$ to $b$ is given by the definite integral:

Substituting the given function and limits:

We integrate term by term:

Using the power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ and $\int c\,dx = cx$:

Let $F(x) = \frac{x^3}{15} + 3x$ be the antiderivative.

According to the Fundamental Theorem of Calculus, the definite integral is $F(b) - F(a)$.
Evaluate $F(5)$:

Simplify $\frac{125}{15}$ by dividing both by 5: $\frac{25}{3}$.

Evaluate $F(2)$:

Combine the constant terms and the fractional terms:

Simplify $\frac{117}{15}$ by dividing both by 3: $\frac{39}{5}$.

Convert 9 to a fraction with denominator 5: $9 = \frac{45}{5}$.

The area under the graph is $\frac{84}{5}$.