30. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis

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30. Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and x-axis

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Vishal Baghel | Contributor-Level 10

4 months ago

The given equation of the parabola is $x^{2} = y$ ---------(1)

and that the line is $y = x + 2$ --------------(2)

Solving (1) and (2) for x and y

$\begin{array}{l} x^{2} = x + 2 \\ \to x^{2} - x - 2 = 0 \\ = x^{2} + x - 2 x - 2 = 0 \\ = x (x + 1) - 2 (x + 1) = 0 \\ = (x + 1) (x - 2) = 0 \\ = x = - 1 & x = 2 \end{array}$

When $x = - 1, y = {(- 1)}^{2} = 1$

And $x = 2, y = 2^{2} = 4$

$∴$ The point of intersection of the parabola and the lines $A (- 1, 1) & B (2, 4)$

Hence the required area enclosed region is $e a (A O B A) = a r e a (D A B C D) - a r e a (D A O B C D)$

$\begin{array}{l} = \int_{- 1}^{2} y_{l i n e} d x - \int_{- 1}^{2} y_{c u r v e} d x \\ = \int_{- 1}^{2} (x + 2) d x - \int_{- 1}^{2} x^{2} d x \\ = {[\frac{x^{2}}{2} 2 x]}_{- 1}^{2} - {[\frac{x^{3}}{3}]}_{- 1}^{2} \end{array}$

$\begin{array}{l} = [(\frac{2^{2}}{2} 2.2) - (\frac{{(- 1)}^{2}}{2} + 2 (- 1))] - [\frac{2^{3}}{3} - \frac{{(- 1)}^{3}}{3}] \\ = [2 + 4 - \frac{1}{2} + 2] - [\frac{8 + 1}{3}] = \frac{15}{2} - 3 = \frac{9}{2} u n i t^{2} \end{array}$

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Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

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