The area bounded by the straight lines x = 0, | vedclass

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(D) The required area is given by the integral of the upper curve minus the lower curve between the limits $x = 0$ and $x = 2$.Area = $\int_0^2 [2^x -

Vedclass · Accepted Answer

$\frac{3}{\log 2} - \frac{4}{3}$

Vedclass · Answer

(D) The required area is given by the integral of the upper curve minus the lower curve between the limits $x = 0$ and $x = 2$.Area = $\int_0^2 [2^x - (2x - x^2)] \, dx$$= \int_0^2 (2^x - 2x + x^2) \, dx$$= \left[ \frac{2^x}{\log 2} - x^2 + \frac{x^3}{3} \right]_0^2$$= \left( \frac{2^2}{\log 2} - 2^2 + \frac{2^3}{3} \right) - \left( \frac{2^0}{\log 2} - 0^2 + \frac{0^3}{3} \right)$$= \left( \frac{4}{\log 2} - 4 + \frac{8}{3} \right) - \left( \frac{1}{\log 2} \right)$$= \frac{4}{\log 2} - \frac{1}{\log 2} - 4 + \frac{8}{3}$$= \frac{3}{\log 2} - \frac{12}{3} + \frac{8}{3}$$= \frac{3}{\log 2} - \frac{4}{3}$.

The area bounded by the straight lines $x = 0, x = 2$ and the curves $y = 2^x, y = 2x - x^2$ is

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