5.5 Integration and area under a curveIB Maths: Applications and Interpretation SL: Revision notes

Integration reverses differentiation. An antiderivative of $f(x)$ is a function whose derivative is $f(x)$. For $n\neq-1$: $\int ax^n\,dx=\frac{a\,x^{n+1}}{n+1}+c.$ Raise the power by one, then divide by the new power. Integrate a sum term by term. A constant $k$ integrates to $kx$. Negative powers work too: $\int x^{-2}\,dx=-x^{-1}+c=-\frac1x+c$. Example: $\int(6x^2-4x+3)\,dx=2x^3-2x^2+3x+c$. Check by differentiating: you must get back $6x^2-4x+3$.

If you are given a point on the curve, or a value of $y$ for a particular $x$, you can find $c$. Integrate, then substitute the values and solve for $c$. Example: $\frac{dy}{dx}=3x^2+x$ and $y=10$ when $x=1$. $y=x^3+\frac12x^2+c$. Substituting: $10=1+\frac12+c$, so $c=8.5$ and $y=x^3+\frac12x^2+8.5$. Once you have $c$, you can find $y$ for other values of $x$.

A definite integral has limits: $\int_a^b f(x)\,dx=F(b)-F(a),$ where $F$ is an antiderivative of $f$. The constant $c$ cancels, so it is not needed. Example: $\int_2^6(3x^2+4)\,dx=\left[x^3+4x\right]_2^6=(216+24)-(8+8)=224$. On your GDC, enter the function and the limits to evaluate a definite integral directly. Always subtract in the order 'upper limit minus lower limit'. Reversing the limits changes the sign of the answer.

If $f(x)>0$ for $a\le x\le b$, the area of the region enclosed by the curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$ is $\text{Area}=\int_a^b f(x)\,dx.$ Write this expression first, with the correct limits, and then evaluate it. Example: the area under $y=3x^2+4$ from $x=2$ to $x=6$ is $\int_2^6(3x^2+4)\,dx=224$ square units. So there is a link between antiderivatives, definite integrals and area: the definite integral of a positive function is the area under its graph.

Use the GDC to evaluate definite integrals, and use graphing or dynamic geometry software to see the region. Give areas to three significant figures unless exact values are required, with units such as m$^2$. In context, the integral can represent an area (for example the cross-section of a tunnel). Interpret answers sensibly, and compare areas by comparing the integrals, for example a percentage of a whole.