When the graphs of the functions $\sin x$sinx and $\cos x$cosx are plotted, we see that there is a sequence of regions bounded by the two graphs. There is one such a bounded region between $x=0$x=0 and $x=2\pi$x=2π.
To find its area, we need to know exactly where the region begins and ends. That is, we need to know where the two graphs intersect. We solve the equation $\sin x=\cos x$sinx=cosx.
In this example, we may recognise that the equation is satisfied when $x=\frac{\pi}{4}$x=π4 and when $x=\frac{5\pi}{4}$x=5π4. If not, we might manipulate the equation by dividing both sides by $\cos x$cosx to obtain $\tan x=1$tanx=1 and proceed from there. The diagram is shown below.
We carry out an integration to find the area of the region bounded by the sine and cosine curves between $\frac{\pi}{4}$π4 and $\frac{5\pi}{4}$5π4.
Observe that on the domain $\frac{\pi}{4}
The integral to be evaluated is $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\ \left(\sin x-\cos x\right)\ \mathrm{d}x$∫5π4π4 (sinx−cosx) dx.
The antiderivatives are well-known. So, we can write
$I$I | $=$= | $\left[-\cos x-\sin x\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$[−cosx−sinx]5π4π4 |
$=$= | $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)$1√2+1√2−(−1√2−1√2) | |
$=$= | $\frac{4}{\sqrt{2}}$4√2 | |
$=$= | $2\sqrt{2}$2√2 |
Find the area of the enclosed region between $f(x)=\sin x+1$f(x)=sinx+1 and $g(x)=\cos x$g(x)=cosx. These functions intersect when $\sin x+1=\cos x$sinx+1=cosx. By inspection, this occurs at $x=0$x=0 and at $x=\frac{3\pi}{2}$x=3π2.
From the graph, it is clear that $\sin x+1$sinx+1 is always greater than $\cos x$cosx on the interval $\left(0,\frac{3\pi}{2}\right)$(0,3π2).
The integral to be evaluated is $\int_0^{\frac{3\pi}{2}}\ \left(\sin x+1-\cos x\right)\ \mathrm{d}x$∫3π20 (sinx+1−cosx) dx.
This is
$\left[-\cos x+x-\sin x\right]_0^{\frac{3\pi}{2}}$[−cosx+x−sinx]3π20
$=0+\frac{3\pi}{2}-0+1-0+0=\frac{5\pi}{2}$=0+3π2−0+1−0+0=5π2.
In problems of areas of enclosed regions between sine and cosine curves, it may not always be easy to find the endpoints of the interval over which an integral will be evaluated. Consider the functions $2\cos(2x)$2cos(2x) and $\sin x+1$sinx+1.
We need to solve the equation $2\cos(2x)=\sin x+1$2cos(2x)=sinx+1. Some trigonometric identities may be needed.
$2\cos(2x)$2cos(2x) | $=$= | $\sin x+1$sinx+1 |
$2(1-2\sin^2x)$2(1−2sin2x) | $=$= | $\sin x+1$sinx+1 |
$-4\sin^2x-\sin x+1$−4sin2x−sinx+1 | $=$= | $0$0 |
This is a quadratic in $\sin x$sinx. Its solution, given by the quadratic formula is, $\sin x=\frac{1\pm\sqrt{1+16}}{-8}$sinx=1±√1+16−8. So, $\sin x\approx-0.64$sinx≈−0.64 or $\sin x\approx0.39$sinx≈0.39.
The inverse sine function on a calculator then gives solutions $x=-0.6945$x=−0.6945, which is about $-0.2211\pi$−0.2211π, and $x=0.4006$x=0.4006, which is about $0.1275\pi$0.1275π.
Thus, between $x=0$x=0 and $x=2\pi$x=2π, we should find intersection points at $0.1275\pi$0.1275π, $0.8725\pi$0.8725π, $1.2211\pi$1.2211π and $1.7789\pi$1.7789π. The intersection points are shown on the graph below.
We could find the area enclosed between $0.1275\pi$0.1275π and $0.8725\pi$0.8725π or the area between $0.8725\pi$0.8725π and $1.2211\pi$1.2211π, or the area between $1.2211\pi$1.2211π and $1.7789\pi$1.7789π.
As an exercise, we calculate the enclosed area between $0.8725\pi$0.8725π and $1.2211\pi$1.2211π. In this region, $2\cos(2x)>\sin x+1$2cos(2x)>sinx+1. So, we evaluate
$I$I | $=$= | $\int_{0.8725\pi}^{1.2211\pi}\ \left(2\cos(2x)-(\sin x+1)\right)\ \mathrm{d}x$∫1.2211π0.8725π (2cos(2x)−(sinx+1)) dx |
$=$= | $\left[\sin(2x)+\cos x-x\right]_{0.8725\pi}^{1.2211\pi}$[sin(2x)+cosx−x]1.2211π0.8725π | |
$=$= | $-3.6209+4.38$−3.6209+4.38 | |
$=$= | $0.7591$0.7591 |
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