5.11 Definite integrals and areasIB Maths: Analysis and Approaches SL: Revision notes

A definite integral has limits and gives a number, not a function. If $g$ is any antiderivative of the integrand, then $\int_a^b g'(x)\,\mathrm{d}x = g(b) - g(a).$ The constant of integration cancels, so you can leave it out. For example $\int_1^2 (3x^2 + 2x)\,\mathrm{d}x = \left[x^3 + x^2\right]_1^2 = 12 - 2 = 10.$ Useful properties: $\int_a^b k f(x)\,\mathrm{d}x = k\int_a^b f(x)\,\mathrm{d}x$, $\int_a^b (f+g)\,\mathrm{d}x = \int_a^b f\,\mathrm{d}x + \int_a^b g\,\mathrm{d}x$ and $\int_a^a f(x)\,\mathrm{d}x = 0$.

Because $\int_a^b h'(x)\,\mathrm{d}x = h(b) - h(a)$, a definite integral of a rate gives the change in the quantity. If you know $h(a)$ you can find $h(b) = h(a) + \int_a^b h'(x)\,\mathrm{d}x.$ For example, if $h'(x) = \frac{4}{2x+1}$ and $h(0) = 3$, then $h(4) = 3 + \left[2\ln(2x+1)\right]_0^4 = 3 + 2\ln 9$.

Some definite integrals cannot be done by hand (for example $\int_0^1 e^{-x^2}\,\mathrm{d}x$). On Paper 2 these are found with the GDC; write the integral down first, then give the value to 3 s.f.

If $f(x) \ge 0$ on $[a,b]$, the area between $y = f(x)$ and the $x$-axis is $\int_a^b f(x)\,\mathrm{d}x$.

If the curve goes below the axis, the integral over that part is negative. So:

1. Find where the curve crosses the $x$-axis (solve $f(x) = 0$).
2. Integrate separately over each interval.
3. Add the absolute values.

For $y = x^3 - 4x$ between $x=-2$ and $x=2$: $\int_{-2}^{0} = 4$ and $\int_{0}^{2} = -4$, so the area is $8$, even though $\int_{-2}^{2}(x^3-4x)\,\mathrm{d}x = 0$. On Paper 1 you must do this without technology.

In context, the sign of an integral has meaning. If $r(t) = 6t - t^2$ is a rate of flow into a reservoir for $0 \le t \le 8$:

• $\int_0^8 r(t)\,\mathrm{d}t = \frac{64}{3}$ is the net change in volume (in minus out).
• Since $r(t) < 0$ for $6 < t \le 8$, the total flowing out is $\left|\int_6^8 r(t)\,\mathrm{d}t\right| = \frac{44}{3}$, and the total flowing in is $\int_0^6 r(t)\,\mathrm{d}t = 36$.

Check: $36 - \frac{44}{3} = \frac{64}{3}$.

The area between $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a,b]$, is $\int_a^b \left(f(x) - g(x)\right)\mathrm{d}x.$ Steps:

1. Solve $f(x) = g(x)$ to find the intersection points; these are usually the limits.
2. Decide which curve is on top in each interval (test a value).
3. Write the integral expression first, then evaluate. The IB expects to see the integral.

It does not matter if the curves are below the $x$-axis: top minus bottom is always correct. For $y = 5x$ and $y = x^3 - 4x$, which meet at $x = 0, \pm 3$, each region has area $\int_0^3 (9x - x^3)\,\mathrm{d}x = \frac{81}{4}$, total $\frac{81}{2}$.