Areas and the definite integral

The definite integral as the area under a curve, evaluating definite integrals with area formulae for triangles, trapezia and circles, and trapping an integral between upper and lower rectangle sums.

The definite integral

For a curve $y = f(x)$ that is above the $x$-axis from $x = a$ to $x = b$, the definite integral $$\int_a^b f(x)\,dx$$ is the area between the curve, the $x$-axis and the vertical lines $x = a$ and $x = b$. The numbers $a$ and $b$ are the lower and upper limits, and $f(x)$ is the integrand.

The area under y = x²

The shaded area is the integral of x² from 0 to b. You will see how to calculate it in the next section.

Try this

Set $b = 3$. Is the area more or less than half of the $3 \times 9$ rectangle?

Using area formulae

Example 1: a triangle●

Evaluate $\displaystyle\int_0^3 2x\,dx$.

The region is a triangle with base $3$ and height $2 \times 3 = 6$.

Area $= \tfrac12 \times 3 \times 6 = 9$.

Example 2: a trapezium●

Evaluate $\displaystyle\int_1^4 (x + 1)\,dx$.

The parallel sides are $f(1) = 2$ and $f(4) = 5$, and the width is $3$.

Area $= \tfrac12(2 + 5) \times 3 = 10.5$.

Example 3: a semicircle●●

Evaluate $\displaystyle\int_{-2}^{2} \sqrt{4 - x^2}\,dx$.

$y = \sqrt{4 - x^2}$ is the upper half of the circle $x^2 + y^2 = 4$.

Area $= \tfrac12 \pi \times 2^2 = 2\pi$.

Trapping an integral

When no formula fits, split the interval into strips. Rectangles that sit under the curve give a lower sum, and rectangles that reach above it give an upper sum. The integral is between the two.

Example 4●●

Use two strips of width $1$ to find bounds for $\displaystyle\int_0^2 x^2\,dx$.

$y = x^2$ is increasing on $0 \le x \le 2$, so the left end of each strip gives the lowest height and the right end the highest.

Lower sum $= 1 \times 0 + 1 \times 1 = 1$. Upper sum $= 1 \times 1 + 1 \times 4 = 5$.

So $1 < \displaystyle\int_0^2 x^2\,dx < 5$. (The exact value turns out to be $\tfrac83$.)

Flashcards

$displaystyleint_a^b f(x),dx$

The area under $y = f(x)$ from $x = a$ to $x = b$ (when the curve is above the axis).

Integrand

The function being integrated.

Lower and upper sums

Rectangles under and over the curve. The integral lies between them.

Practice

Evaluate $\displaystyle\int_0^3 2x\,dx$ using an area formula.
Evaluate $\displaystyle\int_1^4 (x + 1)\,dx$.
$\displaystyle\int_{-2}^{2} \sqrt{4 - x^2}\,dx = $
Evaluate $\displaystyle\int_0^5 3\,dx$.
Evaluate $\displaystyle\int_0^4 (4 - x)\,dx$.
Using two strips of width $1$, find the lower sum for $\displaystyle\int_0^2 x^2\,dx$.
Find the upper sum for the same integral.
$\displaystyle\int_{-3}^{3} \sqrt{9 - x^2}\,dx = $
In $\displaystyle\int_1^5 x^3\,dx$, the upper limit is: