Definite integrals: worked questions

This lesson evaluates a series of definite integrals. The key idea is the fundamental theorem of calculus: if $F$ is an antiderivative of $f$, then

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a).$

Unlike indefinite integrals, a definite integral has no constant of integration.

A polynomial

Evaluate $\displaystyle\int_1^2 (x^2 + 2x + 5)\,dx$. Integrate term by term:

$\int_1^2 (x^2 + 2x + 5)\,dx = \left[\frac{x^3}{3} + x^2 + 5x\right]_1^2.$

Upper limit $x=2$: $\tfrac{8}{3} + 4 + 10 = \tfrac{8}{3} + 14$.

Lower limit $x=1$: $\tfrac{1}{3} + 1 + 5 = \tfrac{1}{3} + 6$.

Subtracting,

$\left(\frac{8}{3} + 14\right) - \left(\frac{1}{3} + 6\right) = \frac{8 + 42}{3} - \frac{1 + 18}{3} = \frac{50}{3} - \frac{19}{3} = \frac{31}{3}.$

The reciprocal function

Evaluate $\displaystyle\int_1^3 \frac{1}{x}\,dx$. Since $\int \tfrac{1}{x}\,dx = \ln|x|$,

$\int_1^3 \frac{1}{x}\,dx = \big[\ln|x|\big]_1^3 = \ln 3 - \ln 1 = \ln 3 - 0 = \ln 3.$

An exponential

Evaluate $\displaystyle\int_4^5 a^x\,dx$. Using $\int a^x\,dx = \dfrac{a^x}{\ln a}$,

$\int_4^5 a^x\,dx = \left[\frac{a^x}{\ln a}\right]_4^5 = \frac{a^5 - a^4}{\ln a}.$

Using a standard formula

Evaluate $\displaystyle\int_2^3 \frac{dx}{x^2 - 1}$. Here $x^2 - 1 = x^2 - 1^2$, so use

$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\,\ln\left|\frac{x - a}{x + a}\right|,$

with $a = 1$:

$\int_2^3 \frac{dx}{x^2 - 1} = \frac{1}{2}\left[\ln\left|\frac{x - 1}{x + 1}\right|\right]_2^3.$

Upper limit $x=3$: $\ln\dfrac{2}{4} = \ln\dfrac{1}{2}$. Lower limit $x=2$: $\ln\dfrac{1}{3}$. So

$\frac{1}{2}\left(\ln\frac{1}{2} - \ln\frac{1}{3}\right) = \frac{1}{2}\ln\frac{1/2}{1/3} = \frac{1}{2}\ln\frac{3}{2}.$

A quotient by partial fractions

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

$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2},$

so $x = A(x+2) + B(x+1)$.

Find $B$ (put $x = -2$): $-2 = B(-1)$, hence $B = 2$.

Find $A$ (put $x = -1$): $-1 = A(1)$, hence $A = -1$.

Therefore

$\int_1^2 \frac{x}{(x+1)(x+2)}\,dx = \int_1^2 \left(\frac{-1}{x+1} + \frac{2}{x+2}\right)dx = \big[-\ln|x+1| + 2\ln|x+2|\big]_1^2.$

Applying the limits,

$(-\ln 3 + 2\ln 4) - (-\ln 2 + 2\ln 3) = -\ln 3 + \ln 16 + \ln 2 - \ln 9.$

Collecting, $\ln(2 \cdot 16) - \ln(3 \cdot 9) = \ln 32 - \ln 27 = \ln\dfrac{32}{27}.$

Integration by parts

Evaluate $\displaystyle\int_0^1 x\,e^x\,dx$. By parts, taking $x$ as the first function and $e^x$ as the second,

$\int x\,e^x\,dx = x\,e^x - \int 1 \cdot e^x\,dx = x\,e^x - e^x = e^x(x - 1).$

Applying the limits,

$\int_0^1 x\,e^x\,dx = \big[e^x(x - 1)\big]_0^1 = e^1(1 - 1) - e^0(0 - 1) = 0 - (-1) = 1.$

By substitution

Evaluate $\displaystyle\int_1^2 \frac{x}{x^2 + 1}\,dx$. Let $t = x^2 + 1$, so $dt = 2x\,dx$, giving $x\,dx = \tfrac{1}{2}\,dt$.

Method 1: resubstitute, then apply limits

$\int \frac{1}{2}\cdot\frac{dt}{t} = \frac{1}{2}\ln|t| = \frac{1}{2}\ln(x^2 + 1).$

Now apply the $x$ limits $1$ to $2$:

$\frac{1}{2}\big[\ln(x^2 + 1)\big]_1^2 = \frac{1}{2}(\ln 5 - \ln 2) = \frac{1}{2}\ln\frac{5}{2}.$

Method 2: change the limits to $t$

When $x = 1$, $t = 1^2 + 1 = 2$; when $x = 2$, $t = 2^2 + 1 = 5$. Then

$\int_2^5 \frac{1}{2}\cdot\frac{dt}{t} = \frac{1}{2}\big[\ln|t|\big]_2^5 = \frac{1}{2}(\ln 5 - \ln 2) = \frac{1}{2}\ln\frac{5}{2}.$

Both methods agree.

Key takeaways

• Evaluate a definite integral as $F(b) - F(a)$, applying the upper limit minus the lower limit; there is no constant of integration.
• Choose the right technique: a standard formula for $\tfrac{1}{x^2 - a^2}$, partial fractions for a product in the denominator, parts for a product like $x\,e^x$, and substitution for $\tfrac{x}{x^2+1}$.
• With substitution you can either resubstitute back to $x$ before applying the original limits, or change the limits to the new variable; the answer is the same.