The area bounded by the curve y = x^2 + 2,the | vedclass

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(C) The area bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the integral $\int_{a}^{b} y \, dx$.Here,$y =

Vedclass · Accepted Answer

$\frac{13}{3} 	ext{ sq unit}$

Vedclass · Answer

(C) The area bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the integral $\int_{a}^{b} y \, dx$.Here,$y = x^2 + 2$,$a = 1$,and $b = 2$.Therefore,the required area is:$	ext{Area} = \int_{1}^{2} (x^2 + 2) \, dx$$= \left[ \frac{x^3}{3} + 2x ight]_{1}^{2}$$= \left( \frac{2^3}{3} + 2(2) ight) - \left( \frac{1^3}{3} + 2(1) ight)$$= \left( \frac{8}{3} + 4 ight) - \left( \frac{1}{3} + 2 ight)$$= \left( \frac{8 + 12}{3} ight) - \left( \frac{1 + 6}{3} ight)$$= \frac{20}{3} - \frac{7}{3}$$= \frac{13}{3} 	ext{ sq unit}$.

The area bounded by the curve $y = x^2 + 2$,the $x$-axis,and the lines $x = 1$ and $x = 2$ is:

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