Definite integrals and area under a curveEdexcel International A Level Maths: Revision notes

A definite integral has limits: $\int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)$, where $F$ is an antiderivative of $f$. The constant $+c$ is not needed because it cancels. Write the integrand in index form first (e.g. $\sqrt{x}=x^{\frac12}$, $\frac{1}{x^2}=x^{-2}$), then use $\int x^n\,dx=\frac{x^{n+1}}{n+1}$ for $n\ne-1$. Example: $\int_0^2(3x^2-2x+1)\,dx=\left[x^3-x^2+x\right]_0^2=8-4+2=6$.

If $y\ge0$ for $a\le x\le b$, the area between the curve, the $x$-axis and the lines $x=a$ and $x=b$ is $\int_a^b y\,dx$. The definite integral is the limit of the sum of thin strips of height $y$ and width $\delta x$. Example: $y=3\sqrt{x}+\frac{2}{x^2}$ is above the axis for $x>0$. Area from $x=1$ to $x=4$ is $\left[2x^{\frac32}-\frac2x\right]_1^4=\left(16-\frac12\right)-0=\frac{31}{2}$. Integrals of the form $\int x\,dy$ (area measured against the $y$-axis) are not required.

Where the curve is below the $x$-axis, $y<0$ and the integral is negative. The area is the modulus of that integral. If the curve crosses the axis between the limits, split the integral at the root, find each part, and add the magnitudes. Example: $y=x^2-4x+3$ crosses the axis at $x=1$ and $x=3$. $\int_1^3y\,dx=-\frac43$, so the area enclosed is $\frac43$. Between $x=0$ and $x=3$: $\int_0^1y\,dx=\frac43$, $\int_1^3y\,dx=-\frac43$, so the total area is $\frac83$, even though $\int_0^3y\,dx=0$.

To find a region bounded by a curve and given straight lines, identify the limits from the lines (e.g. $x=a$, $x=b$) or from where the curve meets the axis (solve $y=0$). Sketch the curve to see which parts lie above or below the axis. Example: the area under $y=6x-x^2$ and above the axis: roots at $x=0$ and $x=6$, so area $=\int_0^6(6x-x^2)\,dx=\left[3x^2-\frac{x^3}{3}\right]_0^6=108-72=36$. For a region under a line and a curve, a trapezium or triangle can sometimes give the line's area, but integration is always valid.

Find the total area enclosed by $y=x^2-4$, the $x$-axis and the lines $x=0$ and $x=3$. Roots: $x=\pm2$; only $x=2$ lies in the interval. $F(x)=\frac{x^3}{3}-4x$. $\int_0^2y\,dx=F(2)-F(0)=\frac83-8=-\frac{16}{3}$, below the axis, area $\frac{16}{3}$. $\int_2^3y\,dx=F(3)-F(2)=-3+\frac{16}{3}=\frac73$, above the axis, area $\frac73$. Total area $=\frac{16}{3}+\frac73=\frac{23}{3}$. (The signed total $\int_0^3y\,dx=-3$ would be wrong.)