The fundamental theorem of calculus

Evaluating definite integrals with primitives using the fundamental theorem of calculus, including expanding brackets, splitting fractions and negative powers, and the warning about integrating across an asymptote.

The theorem

Fundamental theorem of calculus

$$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a), \quad \text{where } F'(x) = f(x)$$

Any primitive works, because the constant $C$ cancels when you subtract. So leave it out.

Example 1●

Evaluate $\displaystyle\int_1^3 x^2\,dx$.

$\Big[\dfrac{x^3}{3}\Big]_1^3 = 9 - \dfrac13 = \dfrac{26}{3}$

Example 2●

Evaluate $\displaystyle\int_0^2 (3x^2 - 4x + 1)\,dx$.

$\Big[x^3 - 2x^2 + x\Big]_0^2 = (8 - 8 + 2) - 0 = 2$

Preparing the integrand

Example 3: expand first●●

Evaluate $\displaystyle\int_1^2 (2x - 1)^2\,dx$.

$(2x - 1)^2 = 4x^2 - 4x + 1$.

$\Big[\tfrac43 x^3 - 2x^2 + x\Big]_1^2 = \left(\tfrac{32}{3} - 6\right) - \left(\tfrac43 - 1\right) = \tfrac{14}{3} - \tfrac13 = \tfrac{13}{3}$

Example 4: split the fraction●●

Evaluate $\displaystyle\int_1^2 \frac{x^3 + 1}{x^2}\,dx$.

$\dfrac{x^3 + 1}{x^2} = x + x^{-2}$.

$\Big[\dfrac{x^2}{2} - \dfrac1x\Big]_1^2 = \left(2 - \tfrac12\right) - \left(\tfrac12 - 1\right) = 2$

Flashcards

Fundamental theorem

$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$

Brackets in the integrand

Expand first.

A fraction over $x^n$

Split into separate powers of $x$.

Practice

Evaluate $\displaystyle\int_1^3 x^2\,dx$.
Evaluate $\displaystyle\int_0^2 (3x^2 - 4x + 1)\,dx$.
Evaluate $\displaystyle\int_1^2 (2x - 1)^2\,dx$.
Evaluate $\displaystyle\int_1^2 \frac{x^3 + 1}{x^2}\,dx$.
Evaluate $\displaystyle\int_0^1 x^4\,dx$.
Evaluate $\displaystyle\int_1^4 \sqrt{x}\,dx$.
Evaluate $\displaystyle\int_{-1}^{2} (2x + 1)\,dx$.
Why is $\displaystyle\int_{-1}^{1} \frac{1}{x^2}\,dx = -2$ wrong?
Why can the $+\,C$ be left out of a definite integral?