The areas between non-negative sections of the graphs of trigonometric functions and the horizontal axis are found as definite integrals, just as they are with curves generated by other functions.
To find the area enclosed by the cosine curve, and the horizontal axis between $0$0 and $\frac{\pi}{2}$π2, we calculate $\int_0^{\frac{\pi}{2}}\cos t\ \mathrm{d}t$∫π20cost dt.
This is $\sin(\frac{\pi}{2})-\sin0=1$sin(π2)−sin0=1.
The area found by integration may represent the real physical area of a plane shape or it may represent an abstract quantity such as the work done by a variable force. So, depending on the application, it may be necessary to consider positive and negative intervals of a function separately because areas below the horizontal axis will turn out to be negative in the calculation.
To find the area of the shape defined by the sine curve, and the horizontal axis between $0$0 and $2\pi$2π we note that the area found by the integral $\int_0^{2\pi}\sin t\ \mathrm{d}t=-\cos2\pi+\cos0=0$∫2π0sint dt=−cos2π+cos0=0. And yet the shape defined by the graph does not have zero area.
We, instead, consider the positive and negative portions separately and take the absolute value of the negative part. We calculate,
$\int_0^{\pi}\sin t\ \mathrm{d}t+\ \left|\int_{\pi}^{2\pi}\sin t\ \mathrm{d}t\ \right|$∫π0sint dt+ |∫2ππsint dt |.
This is, $-\cos\pi+\cos0-\left(-\cos2\pi+\cos\pi\right)=4$−cosπ+cos0−(−cos2π+cosπ)=4.
We can find the areas of shapes formed between two curves. Roughly speaking, the area between two curves is the positive difference between the area below one of them and the area below the other.
Complications can arise if the curves intersect, in which case the same sort of care is needed as when a curve crosses the horizontal axis.
The graphs of $y=\sin x$y=sinx and $y=1+\sin x$y=1+sinx, and the lines $x=0$x=0 and $x=2\pi$x=2π define a plane shape. What is its area?
The area below $1+\sin x$1+sinx is $\int_0^{2\pi}1+\sin t\ \mathrm{d}t=\left[t-\cos t\right]_0^{2\pi}=2\pi$∫2π01+sint dt=[t−cost]2π0=2π.
From this we need to subtract the area below $\sin x$sinx between $0$0 and $\pi$π, and we need to add the area between $\sin x$sinx and the axis between $\pi$π and$2\pi$2π. These cancel each other and the net result is just $2\pi$2π.
A more efficient way to obtain this result is to reason that we should integrate the difference between the two functions over the required interval. That is, we require
$\int_0^{2\pi}\left(1+\sin t-\sin t\right)\ \mathrm{d}t$∫2π0(1+sint−sint) dt | $=$= | $\int_0^{2\pi}1\ \mathrm{d}t$∫2π01 dt |
$=$= | $\left[t\right]_0^{2\pi}$[t]2π0 | |
$=$= | $2\pi$2π |
Find the area of the shape bounded by the graphs $x=0$x=0, $y=\tan x$y=tanx and $y=1+\sin x$y=1+sinx.
We need to know at what $x$x-value the graphs of $1+\sin x$1+sinx and $\tan x$tanx intersect. Therefore, we must solve $1+\sin x=\tan x$1+sinx=tanx. From the diagram above, the intersection point appears to be at approximately $x=1.1$x=1.1. While there are various means of finding a more accurate approximation, the exact value cannot be written down. We will use $x=1.0826$x=1.0826.
Now, we integrate:
$\int_0^{1.0826}\left(1+\sin t-\tan t\right)\ \mathrm{d}t$∫1.08260(1+sint−tant) dt | $=$= | $\left[t-\cos t+\ln(\cos t)\right]_0^{1.0826}$[t−cost+ln(cost)]1.08260 |
$=$= | $1.0826-0.469-0.7571-0+1-0$1.0826−0.469−0.7571−0+1−0 | |
$=$= | $0.8565$0.8565 |
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply integration methods in solving problems