Answer:
[tex]\dfrac{30\sqrt{3}+5\pi }{12}\text{ square unit}[/tex]
Step-by-step explanation:
Given: Two curve
[tex]y=5\cos(4x)[/tex]
[tex]y=5-5cos(4x)[/tex]
We are given two curve and find area under both curve between [tex]0\leq x\leq \dfrac{\pi}{4}[/tex]
First we draw the graph of both curve on same coordinate plane. Please see the attachment for graph.
Area under the curve [tex]=\int_a^b(y_{\text{Upper}}-y_{\text{Lower}})dx[/tex]
For region 1:
[tex]A_1=\int_0^{\frac{\pi}{12}}(y_1-y_2)dx[/tex]
[tex]A_1=\int_0^{\frac{\pi}{12}}(5\cos(4x)-5+5\cos(4x))dx[/tex]
[tex]A_1=\int_0^{\frac{\pi}{12}}(10\cos(4x)-5)dx[/tex]
[tex]A_1=\dfrac{10}{4}\sin(4x)|_0^{\frac{\pi}{12}}-5x|_0^{\frac{\pi}{12}}[/tex]
[tex]A_1=\dfrac{5\sqrt{3}}{4}-\dfrac{5\pi }{12}\approx 0.8561[/tex]
For region 2:
[tex]A_2=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}(y_2-y_1)dx[/tex]
[tex]A_2=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}(5\cos(4x)-5+5\cos(4x))dx[/tex]
[tex]A_2=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}(5-10\cos(4x))dx[/tex]
[tex]A_2=5x|_{\frac{\pi}{12}}^{\frac{\pi}{4}}-\dfrac{10}{4}\sin(4x)|_{\frac{\pi}{12}}^{\frac{\pi}{4}}[/tex]
[tex]A_2=\dfrac{5\pi }{6}+\dfrac{5\sqrt{3}}{4}\approx 4.7831[/tex]
Total area under the curve, A
[tex]A=A_1+A_2[/tex]
[tex]A=\dfrac{5\sqrt{3}}{4}-\dfrac{5\pi }{12}+\dfrac{5\pi }{6}+\dfrac{5\sqrt{3}}{4}[/tex]
[tex]A=\dfrac{30\sqrt{3}+5\pi }{12}\approx 4.7831[/tex]
Hence, The area of region enclosed by the two curve is [tex]\dfrac{30\sqrt{3}+5\pi }{12}\text{ square unit}[/tex]
