# How do you find the area bounded by y=4^x, y=2^(x+1) and y=2^x+6?

Jan 3, 2017

2.249 areal units.

#### Explanation:

Let

${G}_{1} : y = {4}^{x} = {2}^{2 x}$

${G}_{2} : y = 2 \left({2}^{x}\right)$

${G}_{3} : y = {2}^{x} + 6$

${G}_{2} \mathmr{and} {G}_{3}$ meet at $A \left(\log \frac{6}{\log} 2 , 12\right) = \left(2.585 , 12\right)$, nearly,

using $2 \left({2}^{x}\right) - {2}^{x} + 6$

${G}_{3} \mathmr{and} {G}_{1}$ meet ( after discarding negative ${2}^{x}$)

at $B \left(\log \frac{3}{\log} 2 , 9\right) = \left(1.585 , 9\right)$, nearly,

using ${4}^{x} = {\left({2}^{x}\right)}^{2} = {2}^{x} + 6$.

${G}_{1} \mathmr{and} {G}_{2}$ meet at C ( 1, 4 ), using ${4}^{x} = 2 \left({2}^{x}\right)$.

Note that B is between farther A and nearer C, in the x-direction and is above the curve ${G}_{2} \left(A C\right)$.

Now, the area under reference is the area enclosed by the curved

triangle ABC.

The vertical x = ${x}_{B}$= 1.585 divides the area into two parts

Area = $\int \left({y}_{{G}_{1}} - {y}_{{G}_{2}}\right) \mathrm{dx}$, from x = ${x}_{C} = 1$ to ${x}_{B} = 1.585$

$+ \int \left({y}_{{G}_{3}} - {y}_{{G}_{2}}\right) \mathrm{dx}$, from x = ${x}_{B} = 1.585$ to ${x}_{A} = 2.585$

= int (4^x-2(2^x) dx, for x from 1 to 1.585

$+ \int \left({2}^{x} + 6 - 2 \left({2}^{x}\right)\right) \mathrm{dx}$, for x from 1.585 to 2.585

=[ 4^x/ln 4-2(2^x/ln 2], between x = 1 and 1.585

+[ 2^x/ln2+6x-2(2^x)/ln2]], between x = 1.585 and 2.585

$= \left(3.4924 - 2.8854\right) + \left(6.8536 - 5.1818\right)$

$= 2.279$, nearly..

I would review my answer, a little later, for probable errors.