Question

If $A$ is the area of the figure bounded by the straight lines $x=0$ and $x=2,$ and the curves $y=2^x$ and $y=2x-x^2$ then the value of $672(\frac{3}{\log 2}-A)$ is

Answer 1

The equation $y=2x-x^2$ represents a opening downwards having vertex at $(1,1)$ and crossing the $x$-axis at $(0,0)$ and $(2,0)$.

The equation $y=2^x$ represents the exponential curve as shown in Fig.

Lines $x=0$ and $x=2$ are shown in the figure .

The area bounded by these curves is shaded in the figure.

We slice the shaded region into vertical strips.

For the approxmating rectangle shown in the figure, we have length $(y_1-y_2)$, width $=△x$

$\therefore$ Area $=(y_1-y_2)△x$

The approximating rectangle can move horizontally between $x=0$ and $x=2.$

So, the required area is

$={\int }_0^2(y_1-y_2)dx={\int }_0^2(2^x-2x+x^2)dx$

$\because P(x,y_1)$ and $Q(x,y_2)$ lie on $y=2^x$ and $y=2x-x^2$ respectively.

$\therefore y_1=2^x$ and $y_2=2x-x^2$

$={[\frac{2x}{\log 2}-x^2+\frac{x^3}{3}]}^2$

$=\frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}=(\frac{3}{\log 2}-\frac{4}{3})$ sq. units