Answer:
1.335 (3 d.p.)
Step-by-step explanation:
To find the area enclosed by curves and lines, use definite integration:
[tex]\displaystyle \int^b_a \text{f}(x)\:\:\text{d}x \quad \textsf{(where a is the lower limit and b is the upper limit)}[/tex]
Definite integrals have limits. The limits tell you the range of x-values to integrate the function between.
Given functions:
[tex]\begin{cases}y=\sin x\\y=\cos x \end{cases}[/tex]
Given limits:
As the two curves intersect in the interval 0 ≤ x ≤ 2, the integral needs to be separated into two integrations with the limits of the first integral from x = 0 to the point of intersection, and the limits of the second integral from the point of intersection to x = 2.
Find the point of intersection by equating the functions and solving for x:
[tex]\begin{aligned}\implies \sin x & = \cos x\\\dfrac{\sin x}{\cos x} & = 1\\\tan x & = 1\\x & = \tan^{-1} (1)\\\implies x & = \dfrac{\pi}{4}\end{aligned}[/tex]
Therefore the limits are:
To find the area enclosed by the two curves, integrate the composite function of the bottom curve subtracted from the top curve.
[tex]\boxed{\begin{minipage}{6 cm}\underline{Integrating Trigonometric functions}\\\\$\displaystyle \int sin(x)\:\text{d}x=- \cos (x)+\text{C}\\\\ \int \cos (x)\:\text{d}x=\sin(x)+\text{C}$\\\end{minipage}}[/tex]
Integral 1: R₁
(Area between the curves to the left of the point of intersection)
As the top curve is y=cos(x) and the bottom curve is y=sin(x) between the interval 0 ≤ x ≤ π/4, integrate the composite function g(x)=cos(x)-sin(x):
[tex]\begin{aligned}\displaystyle \int^{\frac{\pi}{4}}_0 \cos x - \sin x \:\: \text{d}x & = \left[\sin x + \cos x\right]^{\frac{\pi}{4}}_0\\& = \left(\sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right)\right) - \left(\sin (0) + \cos (0)\right)\\& = \left(\dfrac{\sqrt{2}}{2} +\dfrac{\sqrt{2}}{2} \right) - (0+ 1)\\& = \sqrt{2}-1\end{aligned}[/tex]
Integral 2: R₂
(Area between the curves to the right of the point of intersection)
As the top curve is y=sin(x) and the bottom curve is y=cos(x) between the interval π/4 ≤ x ≤ 2, integrate the composite function h(x)=sin(x)-cos(x):
[tex]\begin{aligned}\displaystyle \int^2_{\frac{\pi}{4}} \sin x - \cos x \:\: \text{d}x & = \left[-\cos x - \sin x \right]^2_{\frac{\pi}{4}}\\& = \left(- \cos 2 - \sin 2 \right) - \left( - \cos \left (\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{4}\right)\right) \\& = \left(-0.493150...\right) - \left(-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\right)\\& = 0.9210629721...\end{aligned}[/tex]
Finally, add the areas of both regions together to find the total enclosed area:
[tex]\begin{aligned}\implies \textsf{Total Area} & = R_1+R_2 \\& = (\sqrt{2}-1)+0.9210629721...\\& = 1.335276534...\end{aligned}[/tex]
Therefore, the area enclosed by the graphs of y=sin(x), y=cos(x) and x=2 is 1.335 (3 d.p.).
Learn more about definite integrals here:
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