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Application of Integrals

Multiple Choice Questions

111.

The area between the parabola y = x² and the line y = x is :

$\frac{1}{6} sq unit$
$\frac{1}{3} sq unit$
$\frac{1}{2} sq unit$
None of these

112.

The area enclosed between the curves y = x³ and y = $\sqrt{x}$ is, (in square units)

$\frac{5}{3}$
$\frac{5}{4}$
$\frac{5}{12}$
$\frac{12}{5}$

113.

The area included between the parabolas x² = 4y and y² = 4x is (in square units)

$\frac{4}{3}$
$\frac{1}{3}$
$\frac{16}{3}$
$\frac{8}{3}$

114.
The area of the figure bounded by the curves y = cos(x) and y = sin(x) and the coordinate x = 0 and x = $\frac{π}{4}$

$\sqrt{2} + 1$

$\sqrt{2} - 1$

$\frac{1}{\sqrt{2}}$

$\frac{\sqrt{2} - 1}{\sqrt{2}}$

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$\sqrt{2} - 1$

$A_{1} = \int_{0}^{\frac{π}{4}} \cos (x) dx = {[\sin (x)]}_{0}^{\frac{π}{4}} = \frac{1}{\sqrt{2}} A_{2} = \int_{0}^{\frac{π}{4}} \sin (x) dx = {[- \cos (x)]}_{0}^{\frac{π}{4}} = 1 - \frac{1}{\sqrt{2}} A = A_{1} - A_{2} = \frac{1}{\sqrt{2}} - 1 + \frac{1}{\sqrt{2}} = \sqrt{2} - 1$