Areas Under and Between Curves | Inter 2nd Year

Integrals as areas

Ask a farmer the area of an oddly shaped field with one curved boundary and the usual formulas give up. The definite integral does not. In this lesson you will use it to find areas bounded by a curve and the $x$ - or $y$ -axis, areas of parabolic regions, circles and ellipses, and finally the area trapped between two curves.

Area under a curve

If $f(x) \ge 0$ on $[a, b]$ , the area between $y = f(x)$ , the $x$ -axis and the lines $x = a$ , $x = b$ is

$\displaystyle A = \int_a^b y\,dx.$

Two cautions. If the curve lies below the axis, the integral comes out negative, so take its absolute value. If the curve crosses the axis, split the interval at the crossing and add the pieces separately. For area measured against the $y$ -axis, between $y = c$ and $y = d$ , use $\displaystyle \int_c^d x\,dy$ .

Example 1. The area under $y = x^2 + 1$ from $x = 0$ to $x = 3$ is $\displaystyle \left[\tfrac{x^3}{3} + x\right]_0^3 = 12$ .

Example 2. Here is where the caution matters. The area between $y = \sin x$ and the $x$ -axis for $0 \le x \le 2\pi$ : $\displaystyle \int_0^\pi\sin x\,dx + \left\lvert\int_\pi^{2\pi}\sin x\,dx\right\rvert = 2 + 2 = 4$ . The plain integral over $[0, 2\pi]$ would have given $0$ , which is clearly not the area.

Graph of y = sin x from 0 to 2 pi: the hump above the axis from 0 to pi is shaded with area 2 and the hump below the axis from pi to 2 pi with area 2, total 4. — The part below the axis counts as positive area: 2 + 2 = 4.

Example 3. For the region bounded by $y^2 = 9x$ and the line $x = 4$ , use symmetry about the horizontal axis: $\displaystyle 2\int_0^4 3\sqrt{x}\,dx = 6 \cdot \tfrac{2}{3} \cdot 8 = 32$ .

The parabola y squared = 9x opening to the right, cut by the vertical line x = 4 at (4, 6) and (4, -6); the region between them is shaded, area 32. — The region bounded by y squared = 9x and x = 4, symmetric about the x-axis.

Example 4. Now the familiar formula for the circle $x^2 + y^2 = r^2$ drops out: $\displaystyle 4\int_0^r\sqrt{r^2 - x^2}\,dx = 4 \cdot \tfrac{\pi r^2}{4} = \pi r^2$ . In the same way, the ellipse $\displaystyle \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$ has area $\displaystyle 4\int_0^a\tfrac{b}{a}\sqrt{a^2 - x^2}\,dx = \pi ab$ .

Example 5. For the area between $x^2 = 4y$ , the $y$ -axis and the lines $y = 1$ , $y = 4$ in the first quadrant, integrate along the vertical axis: $\displaystyle \int_1^4 2\sqrt{y}\,dy = \tfrac{4}{3}(8 - 1) = \tfrac{28}{3}$ .

Area between two curves

If $f(x) \ge g(x)$ on $[a, b]$ , the area between them is $\displaystyle \int_a^b (f - g)\,dx$ . Always find where the curves meet first; those give you the limits.

Example 6. $y = x^2$ and $y = 2x$ meet at $x = 0, 2$ , and the line is on top, so the area is $\displaystyle \int_0^2(2x - x^2)\,dx = 4 - \tfrac{8}{3} = \tfrac{4}{3}$ .

The parabola y = x squared and the line y = 2x meeting at (0, 0) and (2, 4); the region between them, with the line on top, is shaded and marked 4/3. — Between x = 0 and x = 2 the line lies above the parabola.

Example 7. $y^2 = 4x$ and $x^2 = 4y$ meet at $(0, 0)$ and $(4, 4)$ , giving the area $\displaystyle \int_0^4\left(2\sqrt{x} - \tfrac{x^2}{4}\right)dx = \tfrac{32}{3} - \tfrac{16}{3} = \tfrac{16}{3}$ .

The parabolas y squared = 4x (opening right) and x squared = 4y (opening up) meeting at (0, 0) and (4, 4); the lens-shaped region between them is shaded and marked 16/3. — The two parabolas enclose a lens of area 16/3.

Try these yourself

Find the area under $y = 3x^2$ from $x = 1$ to $x = 2$ .
Find the area bounded by $y = x^3$ , the $x$ -axis and $x = -1$ , $x = 2$ .
Find the area enclosed by $y^2 = 8x$ and the line $x = 2$ .
Find the area of the ellipse $\displaystyle \tfrac{x^2}{25} + \tfrac{y^2}{9} = 1$ .
Find the area bounded by $y = \cos x$ and the $x$ -axis for $0 \le x \le \pi$ .
Find the area between $y = x$ and $y = x^2$ .
Find the area between the parabola $y = 4 - x^2$ and the $x$ -axis.
Find the area bounded by $y = x^2$ and the line $y = x + 2$ .
Find the area of the smaller region cut from the circle $x^2 + y^2 = 16$ by the line $x = 2$ .
Find the area of the triangle with vertices $(0, 0)$ , $(4, 0)$ , $(2, 3)$ by integration.

Answers to check against

Show answers

$\left[x^3\right]_1^2 = 7$ .
$\displaystyle \tfrac{1}{4} + 4 = \tfrac{17}{4}$ .
$\displaystyle 2\int_0^2 2\sqrt{2}\sqrt{x}\,dx = \tfrac{32}{3}$ .
$15\pi$ .
$1 + 1 = 2$ .
$\displaystyle \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$ .
$\displaystyle \int_{-2}^{2}(4 - x^2)\,dx = \tfrac{32}{3}$ .
They meet at $x = -1, 2$ : $\displaystyle \int_{-1}^{2}(x + 2 - x^2)\,dx = \tfrac{9}{2}$ .
$\displaystyle 2\int_2^4\sqrt{16 - x^2}\,dx = \tfrac{16\pi}{3} - 4\sqrt{3}$ .
The sides are the lines $\displaystyle y = \tfrac{3x}{2}$ and $\displaystyle y = 6 - \tfrac{3x}{2}$ : $\displaystyle \int_0^2\tfrac{3x}{2}dx + \int_2^4\left(6 - \tfrac{3x}{2}\right)dx = 3 + 3 = 6$ .

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Areas Under Curves and Between Curves

Integrals as areas

Area under a curve

Area between two curves

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Integrals as areas

Area under a curve

Area between two curves

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