Definite integrals and the area under a curve

An indefinite integral is a family of functions; a definite integral is a single number. Put limits on an integral and it measures the area between a curve and the $x$-axis, which is the first big application of integration and a near-certain question on every Paper 1. This note covers evaluating definite integrals, the area between a curve and the $x$-axis (including regions below the axis, where the integral comes out negative), and areas measured against the $y$-axis. Regions between a curve and a line or two curves are in area between curves.

Evaluating a definite integral

The constant of integration is not needed. If you used $F(x) + c$, you would get $\big(F(b) + c\big) - \big(F(a) + c\big)$, and the $c$ cancels.

For example,

$$\int_1^3 \left(3x^2 - 2x\right) dx = \Big[x^3 - x^2\Big]_1^3 = (27 - 9) - (1 - 1) = 18$$

Swapping the limits changes the sign: $\displaystyle\int_b^a f(x)\, dx = -\int_a^b f(x)\, dx$. And integrals over adjacent intervals add: $\displaystyle\int_a^b + \int_b^c = \int_a^c$.

Why the integral gives the area

Let $A(x)$ be the area under the curve $y = f(x)$ from a fixed starting line up to the vertical line at $x$. Move that line a tiny distance $\delta x$ to the right. The extra area is a thin strip, almost a rectangle of height $f(x)$ and width $\delta x$:

$$\delta A \approx f(x)\,\delta x \quad\Rightarrow\quad \frac{\delta A}{\delta x} \approx f(x)$$

As $\delta x \to 0$ the approximation becomes exact, so $\dfrac{dA}{dx} = f(x)$. The area function is an integral of $f(x)$. The area from $x = a$ to $x = b$ is then $A(b) - A(a)$, which is exactly $\displaystyle\int_a^b f(x)\, dx$.

The shaded region under $y = 4x - x^2$, between $x = 0$ and $x = 4$, has area

$$\int_0^4 \left(4x - x^2\right) dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3}$$

Regions below the axis

Between $x = a$ and $x = b$, if $f(x) < 0$, every strip has negative "height", so the integral is negative. The area itself is positive: take the size of the integral.

When a region lies partly above and partly below the axis, the positive and negative parts cancel in a single integral. To find the total area you must split at the points where the curve crosses the axis, integrate each part separately, and add their sizes.

For $y = x^3 - 4x$ from $0$ to $3$, the curve is below the axis between $0$ and $2$ and above it between $2$ and $3$. The single integral $\displaystyle\int_0^3$ gives $\tfrac{9}{4}$, which is neither area. Example 4 finds the true total, $\tfrac{41}{4}$.

Areas against the y-axis

A region can instead be bounded by the curve, the $y$-axis and horizontal lines $y = c$ and $y = d$. Slice it into horizontal strips of width $x$ and thickness $\delta y$, and the same argument gives

For example, the region between $y = x^2$ (for $x \ge 0$), the $y$-axis, $y = 1$ and $y = 4$ has $x = \sqrt{y}$, so its area is

$$\int_1^4 y^{\frac{1}{2}}\, dy = \left[\tfrac{2}{3}y^{\frac{3}{2}}\right]_1^4 = \tfrac{2}{3}(8 - 1) = \tfrac{14}{3}$$

The region is to the left of the curve, between the horizontal lines $y = 1$ and $y = 4$ and the $y$-axis.

Worked examples

Practice questions

Answers

1. $\left[\dfrac{x^4}{4} + x^2\right]_0^2 = 4 + 4 = 8$.

2. $\Big[2x^{\frac{3}{2}} - x\Big]_1^4 = (16 - 4) - (2 - 1) = 11$.

3. $\Big[-3x^{-2}\Big]_1^2 = -\tfrac{3}{4} - (-3) = \tfrac{9}{4}$.

4. Roots $x = \pm 3$; the curve is above the axis between them. $\displaystyle\int_{-3}^{3} (9 - x^2)\, dx = \left[9x - \tfrac{x^3}{3}\right]_{-3}^{3} = (27 - 9) - (-27 + 9) = 36$.

5. Roots $x = 1, 4$; the curve is below the axis between them. $\displaystyle\int_1^4 (x^2 - 5x + 4)\, dx = \left[\tfrac{x^3}{3} - \tfrac{5x^2}{2} + 4x\right]_1^4 = \left(\tfrac{64}{3} - 40 + 16\right) - \left(\tfrac{1}{3} - \tfrac{5}{2} + 4\right) = -\tfrac{8}{3} - \tfrac{11}{6} = -\tfrac{9}{2}$. The area is $\tfrac{9}{2}$.

6. Let $F(x) = \tfrac{x^4}{4} - \tfrac{4x^3}{3} + \tfrac{3x^2}{2}$. $\displaystyle\int_0^1 = F(1) - F(0) = \tfrac{1}{4} - \tfrac{4}{3} + \tfrac{3}{2} = \tfrac{5}{12}$ (above the axis). $\displaystyle\int_1^3 = F(3) - F(1) = \left(\tfrac{81}{4} - 36 + \tfrac{27}{2}\right) - \tfrac{5}{12} = -\tfrac{9}{4} - \tfrac{5}{12} = -\tfrac{8}{3}$ (below). Total area $= \tfrac{5}{12} + \tfrac{8}{3} = \tfrac{37}{12}$.

7. $x = y^{\frac{1}{3}}$: $\displaystyle\int_1^8 y^{\frac{1}{3}}\, dy = \tfrac{3}{4}(16 - 1) = \tfrac{45}{4}$. The rectangle from $x = 0$ to $2$, $y = 0$ to $8$ (area $16$) is made of: the region under the curve from $x = 1$ to $2$, the region against the $y$-axis ($\tfrac{45}{4}$), and the $1 \times 1$ square at the bottom left. So the area under the curve is $16 - \tfrac{45}{4} - 1 = \tfrac{15}{4}$. Check: $\displaystyle\int_1^2 x^3\, dx = 4 - \tfrac{1}{4} = \tfrac{15}{4}$.

8. $\displaystyle\int_0^2 (4x + 1)^{\frac{1}{2}}\, dx = \left[\dfrac{(4x + 1)^{\frac{3}{2}}}{6}\right]_0^2 = \dfrac{27 - 1}{6} = \dfrac{13}{3}$.

9. The curve meets the axis at $x = 0$ and $x = k$ and is above it between. $\displaystyle\int_0^k (kx - x^2)\, dx = \left[\tfrac{kx^2}{2} - \tfrac{x^3}{3}\right]_0^k = \tfrac{k^3}{2} - \tfrac{k^3}{3} = \tfrac{k^3}{6} = 36$, so $k^3 = 216$ and $k = 6$.