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The area of the region bounded by [tex]y=\frac{4}{\sqrt{1-x^2} }[/tex] , [tex]y=4e^{-x}[/tex] over the interval [0, 1/2] is 0.521 square units.
For given question,
We need to find the area of the region bounded by [tex]y=\frac{4}{\sqrt{1-x^2} }[/tex] , [tex]y=4e^{-x}[/tex] over the interval [0, 1/2]
The region bounded by given curves is as shown below.
The area of this region would be,
[tex]\Rightarrow A=\int\limits^\frac{1}{2} _0 {y-\frac{4}{\sqrt{1-x^{2}}}} \, dx \\\\\Rightarrow A=\int\limits^\frac{1}{2} _0 {4e^{-x}-\frac{4}{\sqrt{1-x^{2}}}} \, dx[/tex]
Now we evaluate above integral.
[tex]\Rightarrow A=4\int\limits^\frac{1}{2} _0 {e^{-x}-\frac{1}{\sqrt{1-x^{2}}}} \, dx\\\\\Rightarrow A=4[\int\limits^\frac{1}{2} _0 {e^{-x}}\,dx -\int\limits^\frac{1}{2} _0{\frac{1}{\sqrt{1-x^{2}}}} \, dx]\\\\\Rightarrow A=4([e^{-x}]_0^{\frac{1}{2} } -[sin^{-1}(x)]_0^{\frac{1}{2} })\\[/tex]
[tex]\Rightarrow A=4([1-\frac{1}{\sqrt{e} } ] -[\frac{\pi}{6} ])\\\\\Rightarrow A=4(1-\frac{1}{\sqrt{e} } -\frac{\pi}{6} )\\\\\Rightarrow A=|4~\times (-0.13013)|\\\\\Rightarrow A=0.521[/tex]
Therefore, the area of the region bounded by [tex]y=\frac{4}{\sqrt{1-x^2} }[/tex] , [tex]y=4e^{-x}[/tex] over the interval [0, 1/2] is 0.521 square units.
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