Question

I: The area bounded by the curves $y=\sin x,y=\cos x$ and $Y$-axis is $\sqrt{2}-1$ sq. units.
II: The area bounded by $y=\cos x,y=x+1,y=0$ is 3/2 sq. units.

Which of the above statement is correct?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II.

Answer 1

The correct option is C Both I and II

$I)$
The required area will be
${\int }_0^{\frac{\pi }{4}}cos\left(x\right)-sin\left(x\right).dx$
$=\left[sin\right(x)+cos(x){]}_0^{\frac{\pi }{4}}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1$
$=\sqrt{2}-1$ sq units.
$II)$
Required area will be
${\int }_{-1}^{0}x+1.dx+{\int }_0^{\frac{\pi }{2}}cos\left(x\right).dx$
$=[\frac{x^2}{2}+x{]}_{-1}^0+[sin\left(x\right){]}_0^{\frac{\pi }{2}}$
$=-(\frac{1}{2}-1)+1$
$=\frac{3}{2}$ sq units.