Question

Find the area enclosed between sin(x) & cos(x) between x = 0 & $x=\frac{\pi }{2}$.

• A. $2\sqrt{2}$
• B. $2$
• C. $2\sqrt{2}-2$
• D. $2\sqrt{2}+2$

Answer 1

The correct option is C $2\sqrt{2}-2$

Let’s look at the graph of sinx and cosx between x = 0 & $x=\frac{\pi }{2}$.

We can see that $x=\frac{\pi }{4}$ is the point where both sin(x) and cos(x) intersect.
From $[0,\frac{\pi }{4}]cos\left(x\right)\ge sin\left(x\right)andfrom[\frac{\pi }{4},\frac{\pi }{2}]sin\left(x\right)\ge cos\left(x\right)$.
Let the enclosed area be = A
$A={\int }_0^{\frac{\pi }{4}}cos\left(x\right)-sin\left(x\right)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}sin\left(x\right)-cos\left(x\right)dxA=\left[sin\right(x)-\{-cos(x\left)\right\}{]}_0^{\frac{\pi }{4}}-[cos\left(x\right)+sin\left(x\right){]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}A=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\left(1\right)-(1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})A=2\sqrt{2}-2$