Q.Evaluate $\displaystyle\int_{1}^{3} (2x + 1)\,dx$.

Find an antiderivative of $2x+1$, then subtract its values at the two limits using the Fundamental Theorem.

$\big[x^2+x\big]_1^3 = (9+3)-(1+1) = 12-2 = 10$.

Apply the Fundamental Theorem: find an antiderivative $F$, then compute $F(3) - F(1)$.

Antiderivative. Using linearity and the power rule, $\displaystyle\int (2x+1)\,dx = 2\cdot\frac{x^2}{2} + x = x^2 + x$. Take $F(x) = x^2 + x$ (no constant needed for a definite integral).

Evaluate at the limits.

Check (dual-solve): the integrand $y = 2x+1$ is a straight line, so the region under it from $x=1$ to $x=3$ is a trapezium. Its parallel sides are the ordinates $f(1) = 2(1)+1 = 3$ and $f(3) = 2(3)+1 = 7$, and its width is $3 - 1 = 2$. Area $= \tfrac{1}{2}(\text{sum of parallel sides})\times\text{width} = \tfrac{1}{2}(3 + 7)(2) = \tfrac{1}{2}(10)(2) = 10$, matching the antiderivative result.

Geometric shortcut: since $2x+1$ is linear and positive on $[1,3]$, the integral is the trapezium area $\tfrac12(f(1)+f(3))(3-1) = \tfrac12(3+7)(2) = 10$.

The frequent errors are writing a $+c$ in the answer (a definite integral has none) and subtracting in the wrong order — always compute $F(\text{upper}) - F(\text{lower})$, i.e. $F(3)-F(1)$, not $F(1)-F(3)$, which would give $-10$.