Question

Evaluate the integral and interpret it as the area of a region

Answer 1

0.90168

Explanation:

We have the following expression:

`\int ^1_(-1)\mleft|4^x-2^x\mright|dx\:`

We first solve the integral and then evaluate it, like this:

`\begin{gathered} \int ^1_(-1)\mleft|4^x-2^x\mright|dx\: =\int ^0_(-1)(-4^x+2^x)dx+\int ^1_0(4^x-2^x)dx \\ \int ^0_(-1)(-4^x+2^x)dx=-\int ^0_(-1)4^xdx+\int ^0_(-1)2^xdx=\mleft[-(4^x)/(\ln\:\:\:\left(4\right))+(2^x)/(\ln\:\:\:\left(2\right))\mright]^0_{^{}-1} \\ \int ^1_0(4^x-2^x)dx=-\int ^1_04^xdx-\int ^1_02^xdx=\mleft[(4^x)/(\ln\:\left(4\right))-(2^x)/(\ln\:\left(2\right))\mright]^1_{^{}0} \end{gathered}`

We evaluate for each interval:

`\begin{gathered} \mleft[-(4^x)/(\ln \mleft(4\mright))+(2^x)/(\ln \mleft(2\mright))\mright]^0_{^{}-1}=-(4^0)/(\ln (4))+(2^0)/(\ln (2))+(4^(-1))/(\ln (4))-(2^(-1))/(\ln (2))=(1)/(8\ln \mleft(2\mright)) \\ \mleft[(4^x)/(\ln \mleft(4\mright))-(2^x)/(\ln \mleft(2\mright))\mright]^1_{^{}0}=(4^1)/(\ln (4))-(2^1)/(\ln (2))-(4^0)/(\ln (4))+(2^0)/(\ln (2))=(1)/(2\ln\left(2\right)) \\ (1)/(2\ln\left(2\right))+(1)/(8\ln\left(2\right))=(5)/(8\ln\left(2\right))\cong0.90168 \end{gathered}`

The value of the integral is 0.90168 and represents the area of the region.