SATHEE: Area Question 8

Area Question 8

Question 8

The area of the region bounded by the curve $y = f (x)$ , the $X$ -axis and the lines $x = a$ and $x = b$ , where $- \infty < a < b < - 2$ , is

(a) $\int_{a}^{b} \frac{x}{3 [{f (x)}^{2} - 1]} d x + b f (b) - a f (a)$

(b) $- \int_{a}^{b} \frac{x}{3 [{f (x)}^{2} - 1]} d x + b f (b) - a f (a)$

(c) $\int_{a}^{b} \frac{x}{3 [{f (x)}^{2} - 1]} d x - b f (b) + a f (a)$

(d) $- \int_{a}^{b} \frac{x}{3 [{f (x)}^{2} - 1]} d x - b f (b) + a f (a)$

Show Answer

Solution:

Given, $y \leq x^{2} + 3 x$

$\Rightarrow y \leq x + {\frac{3}{2}}^{2} - \frac{9}{4} \Rightarrow x + {\frac{3}{2}}^{2} \geq y + \frac{9}{4}$

Since, $0 \leq y \leq 4$ and $0 \leq x \leq 3$

$∴$ The diagram for the given inequalities is

and points of intersection of curves $y = x^{2} + 3 x$ and $y = 4$ are $(1, 4)$ and $(- 4, 4)$

Now required area

$=\int_{0}^{1}\left(x^{2}+3 x\right) d x+\int_{1}^{3} 4 d x=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}{ }{0}^{1}+[4 x]{1}^{3}$

$= \frac{1}{3} + \frac{3}{2} + 4 (3 - 1) = \frac{2 + 9}{6} + 8 = \frac{11}{6} + 8 = \frac{59}{6}$ sq units

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