This note introduces integration as the reverse of differentiation. You will find indefinite integrals of powers of \(x\), \(\sin x\), \(\cos x\), \(e^x\) and \(\tfrac1x\), use a boundary condition to find the constant, and evaluate definite integrals. It also links the definite integral to the area between a curve and the \(x\)-axis.
Key ideas
Anti-differentiation
Integration undoes differentiation. If \(F'(x)=f(x)\), then \(F\) is an antiderivative of \(f\), and we write
\[ \int f(x)\,dx = F(x)+C. \]This is the indefinite integral of \(f\). The constant \(C\) is needed because constants vanish when you differentiate. The functions \(x^2\), \(x^2+5\) and \(x^2-7\) all have the derivative \(2x\), so \(\int 2x\,dx=x^2+C\) must include all of them.
The power rule for integration
To integrate \(ax^n\): add 1 to the power, then divide by the new power.
\[ \int ax^n\,dx=\frac{a\,x^{n+1}}{n+1}+C,\qquad n\ne -1. \]This works for negative integer powers too. For \(n=-1\), the rule fails (it would divide by zero), and the result is a logarithm instead: \(\int\dfrac1x\,dx=\ln|x|+C\).
Other basic integrals
Read the derivatives of \(\sin x\), \(\cos x\) and \(e^x\) backwards. Since \(\dfrac{d}{dx}(-\cos x)=\sin x\), we get \(\int\sin x\,dx=-\cos x+C\). Also \(\int\cos x\,dx=\sin x+C\) and \(\int e^x\,dx=e^x+C\). Here \(x\) is measured in radians for trigonometric functions.
Sums and multiples
Integrate term by term and keep constant multipliers outside, just as in differentiation. Rewrite fractions as powers first: \(\dfrac{4}{x^3}=4x^{-3}\). Expand brackets before integrating.
Finding the constant
If you know a point on the curve, such as \(f(1)=5\), you can find \(C\). Integrate to get \(f(x)\) with \(+C\), then substitute the point and solve for \(C\).
Definite integrals
A definite integral has limits \(a\) and \(b\). Find an antiderivative \(F\), then subtract:
\[ \int_a^b f(x)\,dx=\bigl[F(x)\bigr]_a^b=F(b)-F(a). \]The constant \(C\) cancels, so you leave it out. Swapping the limits changes the sign of the answer. You can also check a definite integral by using the definite integral feature on your GDC.
Area under a curve
If \(f(x)\ge 0\) for \(a\le x\le b\), then \(\int_a^b f(x)\,dx\) is the area between the curve, the \(x\)-axis and the lines \(x=a\) and \(x=b\). If \(f(x)\le 0\) there, the integral is negative, and the area is its absolute value. If the curve crosses the \(x\)-axis between \(a\) and \(b\), split the integral at each crossing and add the positive areas.
Rules for definite integrals
- \(\displaystyle\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx\) for a constant \(k\).
- \(\displaystyle\int_a^b\bigl(f(x)+g(x)\bigr)dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx\).
- \(\displaystyle\int_a^b f(x)\,dx+\int_b^c f(x)\,dx=\int_a^c f(x)\,dx\). This lets you split an interval at a crossing point.
- \(\displaystyle\int_a^a f(x)\,dx=0\).
Why integration gives area
Split the region under a curve into thin strips of width \(\delta x\) and height \(f(x)\). Each strip has area about \(f(x)\,\delta x\). Adding all of them gives a sum, and the integral is the limit of that sum as the strips become thinner. This is what the symbol \(\int\) stands for: a long letter S for sum. It also explains why the integral and the derivative are opposites: the rate at which the area grows as \(b\) increases is the height of the curve at \(b\).
Using a calculator
For a definite integral that is hard to do by hand, enter the function and the limits into the definite integral feature of your GDC. To find the area of a region bounded by a curve and the \(x\)-axis, use the same feature on each section where the curve stays on one side of the axis, and add the positive values. Use the GDC to check hand-worked answers as well.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\displaystyle\int ax^n\,dx=\frac{ax^{n+1}}{n+1}+C\) | Power rule, for integer \(n\ne -1\). |
| \(\displaystyle\int\frac1x\,dx=\ln|x|+C\) | The exceptional case \(n=-1\). |
| \(\displaystyle\int e^x\,dx=e^x+C\) | Exponential function. |
| \(\displaystyle\int\sin x\,dx=-\cos x+C\) | Note the minus sign. |
| \(\displaystyle\int\cos x\,dx=\sin x+C\) | Radians. |
| \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\) | Definite integral; no constant needed. |
| Area | For \(f\ge 0\), area \(=\int_a^b f(x)\,dx\). For \(f\le 0\), area \(=-\int_a^b f(x)\,dx\). |
Worked examples
Example 1 Find \(\displaystyle\int\left(5x^4-\frac{8}{x^3}+\frac3x\right)dx\).
Rewrite each term as a power: \(5x^4-8x^{-3}+3x^{-1}\), where the last term is \(\dfrac3x\).
\(5x^4\to \dfrac{5x^5}{5}=x^5\).
\(-8x^{-3}\to \dfrac{-8x^{-2}}{-2}=4x^{-2}\).
\(\dfrac3x\to 3\ln|x|\).
Add the constant of integration.
Answer: \(x^5+\dfrac{4}{x^2}+3\ln|x|+C\).
Example 2 The gradient of a curve is \(\dfrac{dy}{dx}=6x^2-4x\), and the curve passes through \((1,5)\). Find the equation of the curve.
Integrate: \(y=\dfrac{6x^3}{3}-\dfrac{4x^2}{2}+C=2x^3-2x^2+C\).
Substitute \(x=1\), \(y=5\): \(5=2-2+C\), so \(C=5\).
Answer: \(y=2x^3-2x^2+5\).
Example 3 (a) Evaluate \(\displaystyle\int_1^3(3x^2-2x)\,dx\). (b) Find the area of the region enclosed by \(y=x^2-4x\) and the \(x\)-axis.
(a) An antiderivative is \(x^3-x^2\).
\(\bigl[x^3-x^2\bigr]_1^3=(27-9)-(1-1)=18-0=18\).
(b) Solve \(x^2-4x=0\): \(x(x-4)=0\), so the curve meets the axis at \(x=0\) and \(x=4\).
Between these values the curve is below the axis (for example at \(x=2\), \(y=-4\)).
\(\displaystyle\int_0^4(x^2-4x)\,dx=\left[\frac{x^3}{3}-2x^2\right]_0^4=\left(\frac{64}{3}-32\right)-0=-\frac{32}{3}\).
The integral is negative because the region is below the axis, so the area is the absolute value.
Answer: (a) 18; (b) area \(=\dfrac{32}{3}\approx 10.7\) square units.
Common errors
- Common error: forgetting \(+C\) in an indefinite integral, which makes the answer incomplete.
- Common error: using the power rule on \(\dfrac1x\), giving \(\dfrac{x^0}{0}\). The integral of \(\dfrac1x\) is \(\ln|x|\).
- Common error: writing \(\int\sin x\,dx=\cos x+C\). The correct result is \(-\cos x+C\).
- Common error: subtracting the wrong way in a definite integral. It is always upper limit minus lower limit.
- Common error: quoting a negative area. If the region lies below the axis, take the positive value. If the curve crosses the axis, split the integral first.
Exam tip: after integrating, differentiate your answer in your head. If you do not get the original expression back, you have made a slip.
Check yourself
- Evaluate \(\displaystyle\int_1^2(4x^3-6x^2+2)\,dx\).
Answer
An antiderivative is \(x^4-2x^3+2x\). At 2: \(16-16+4=4\). At 1: \(1-2+2=1\). The value is \(4-1=3\). - Find \(\displaystyle\int(\cos x-3e^x)\,dx\).
Answer
\(\sin x-3e^x+C\). - Given that \(f'(x)=4x-3\) and \(f(2)=5\), find \(f(x)\).
Answer
\(f(x)=2x^2-3x+C\). Then \(f(2)=8-6+C=5\), so \(C=3\). So \(f(x)=2x^2-3x+3\). - Evaluate \(\displaystyle\int_1^e\frac2x\,dx\).
Answer
\(\bigl[2\ln x\bigr]_1^e=2\ln e-2\ln 1=2(1)-0=2\).