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Application of Integrals

Name: Zigya - For The Curious Learner
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Multiple Choice Questions

231.

The measurement ofthe area bounded by the coordinate axes and the curve y = log_e(x), is

1
2
3
$\infty$

232.

The area common to the curves y² = x and x² = y will be

1 sq unit
$\frac{2}{3} sq unit$
$\frac{1}{4} sq unit$
$\frac{1}{3} sq unit$

233.

Area common to the circle x² + y² = 64 and the parabola y² = 12x is equal to

$(\frac{16}{3}) (4 π + \sqrt{3})$
$(\frac{16}{3}) (8 π - \sqrt{3})$
$(\frac{16}{3}) (4 π - \sqrt{3})$
None of these

234.

Area bounded by the curve y = x³, the x-axis and the ordinates x = - 2 and x = 1 is

- 9
$- \frac{15}{4}$
$\frac{15}{4}$
$\frac{17}{4}$

235.

Find the area of the region {(x, y) : x² $\leq y \leq |x|$ }

$\frac{1}{3} sq unit$
$\frac{4}{3} sq unit$
$\frac{2}{3} sq unit$
None of these

236.

Determine the area included between the curve $y =^{2} \cos (x), 0 \leq x \leq \frac{π}{2}$ and the axes.

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{2 π}{3}$
$\frac{π}{4}$

237.

The area of the figure bounded by the curves $y = |x - 1| and y = 3 - |x|$ is

1 sq units
2 sq units
3 sq units
4 sq units

238.

Area of the region bounded by the parabola y = x² and the curve y = $|x|$ is

3
$\frac{1}{3}$
2
$\frac{1}{2}$

239.
The curve, for which the area of the triangle formed by X-axis, the tangent line at any point P and line OP is equal to a, is given by

$y = x - {Cx}^{2}$

$x = Cy \pm \frac{a^{2}}{y}$

$y = Cy \pm \frac{a^{2}}{x}$

None of these

$x = Cy \pm \frac{a^{2}}{y}$

Tangent drawn at any point (x, y) is

$Y - y = \frac{dy}{dx} (X - x) When Y = 0, X = x - y \frac{dx}{dy}$

$∵ Area of ∆ OPQ = a^{2} ∴ |\frac{1}{2} x - y| = a^{2} \Rightarrow |(x - y \frac{dx}{dy}) y| = 2 a^{2} \Rightarrow xy - y^{2} \frac{dx}{dy} = \pm 2 a^{2} \Rightarrow \frac{dx}{dy} - \frac{x}{y} = \pm \frac{2 a^{2}}{y^{2}} Hence, P = - \frac{1}{y} and Q = \pm \frac{2 a^{2}}{y^{2}} ∴ IF = e^{\int Pdy} = e^{- \int \frac{1}{y} dy} = e^{- \log (y)} = e^{\log (\frac{1}{y})} = \frac{1}{y} Hence, required solution is x \times \frac{1}{y} = \int \pm \frac{2 a^{2}}{y^{2}} \times \frac{1}{y} dy \Rightarrow \frac{x}{y} = \pm \frac{2 a^{2} y^{- 2}}{- 2} + C \Rightarrow x = Cy \pm \frac{a^{2}}{y}$