Using our knowledge of integration of exponential and trigonometric functions together with the fundamental theorem of calculus we can evaluate definite integrals and find areas under the curves of such functions.
Keep the following rules in mind as we look at some examples.
| Function $f\left(x\right)$f(x) | Integral $\int f\left(x\right)dx$∫f(x)dx |
|---|---|
| $e^{ax+b}$eax+b | $\frac{1}{a}e^{ax+b}+C$1aeax+b+C |
| $\cos\left(ax+b\right)$cos(ax+b) | $\frac{1}{a}\sin\left(ax+b\right)+C$1asin(ax+b)+C |
| $\sin\left(ax+b\right)$sin(ax+b) | $-\frac{1}{a}\cos\left(ax+b\right)+C$−1acos(ax+b)+C |
Graph symmetry
We can make use of the symmetry properties of functions, including the period of trigonometric functions and whether a function is odd or even to simplify and more easily evaluate integrals and areas bounded by curves.
| Odd function | Even function |
|---|---|
|
Example:
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Example:
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For odd functions: $f\left(-x\right)=-f\left(x\right)$f(−x)=−f(x) and $\int_{-a}^a\ f(x)\ dx=0$∫a−a f(x) dx=0 |
For even functions:$f\left(-x\right)=f\left(x\right)$f(−x)=f(x) and $\int_{-a}^a\ f(x)\ dx=2\int_0^a\ f(x)\ dx$∫a−a f(x) dx=2∫a0 f(x) dx |
| Periodic or other symmetries |
|---|
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Example:
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For the above trigonometric curve we could use the following symmetry properties of this function: $\int_a^c\ f(x)\ dx=0$∫ca f(x) dx=0 and the shaded area is given by $A=2\int_a^b\ f(x)\ dx$A=2∫ba f(x) dx |
Worked examples
Example 1
Find the exact value of 
Think: Use the fact that $\int e^{ax+b}\ dx=\frac{1}{a}e^{ax+b}+C$∫eax+b dx=1aeax+b+C and evaluate the integral at the endpoints.
Do:
Example 2
Find the area bounded by the curve $y=\sin\left(2x\right)$y=sin(2x), the coordinate axes, and the line $x=\frac{3\pi}{2}$x=3π2.
Think: Since we require the area and not simply the integral we will sketch the function and observe which regions of the graph lie below the $x$x-axis. We can then add the integral for regions which lie above the $x$x-axis to the absolute value of those below or use symmetry properties of the graph.
Do:
From the graph we can see that to find the area we could calculate:
Or using the symmetry of the graph we can simplify the calculation to:
Evaluating this we obtain:
Thus the area bounded by the curve $y=\sin\left(2x\right)$y=sin(2x), the coordinate axes, and the line $x=\frac{3\pi}{2}$x=3π2 is $3$3 units2.
Practice questions
Question 1
Find the exact value of $\int_0^2e^{\frac{3}{2}x}dx$∫20e32xdx.
Question 2
Calculate the area enclosed between the curve $y=e^x$y=ex, the coordinate axes, and the line $x=2$x=2.
Question 3
Consider the following.
Given that $y=e^{3x}\left(x-\frac{1}{3}\right)$y=e3x(x−13), determine $y'$y′.
You may use the substitutions $u=e^{3x}$u=e3x and $v=\left(x-\frac{1}{3}\right)$v=(x−13) in your working.
Hence find the exact value of $\int_6^9xe^{3x}dx$∫96xe3xdx.
Question 4
Evaluate $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\left(4\cos x+\cos4x\right)dx$∫π6−π6(4cosx+cos4x)dx.
Question 5
Find the exact area of the shaded regions bounded by the curve $y=3\cos x$y=3cosx.