ANSWER:
0.90168
STEP-BY-STEP EXPLANATION:
We have the following expression:
[tex]\int ^1_{-1}\mleft|4^x-2^x\mright|dx\: [/tex]We first solve the integral and then evaluate it, like this:
[tex]\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[-\frac{4^x}{\ln\:\:\:\left(4\right)}+\frac{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[\frac{4^x}{\ln\:\left(4\right)}-\frac{2^x}{\ln\:\left(2\right)}\mright]^1_{^{}0} \end{gathered}[/tex]We evaluate for each interval:
[tex]\begin{gathered} \mleft[-\frac{4^x}{\ln \mleft(4\mright)}+\frac{2^x}{\ln \mleft(2\mright)}\mright]^0_{^{}-1}=-\frac{4^0}{\ln (4)}+\frac{2^0}{\ln (2)}+\frac{4^{-1}}{\ln (4)}-\frac{2^{-1}}{\ln (2)}=\frac{1}{8\ln \mleft(2\mright)} \\ \mleft[\frac{4^x}{\ln \mleft(4\mright)}-\frac{2^x}{\ln \mleft(2\mright)}\mright]^1_{^{}0}=\frac{4^1}{\ln (4)}-\frac{2^1}{\ln (2)}-\frac{4^0}{\ln (4)}+\frac{2^0}{\ln (2)}=\frac{1}{2\ln\left(2\right)} \\ \frac{1}{2\ln\left(2\right)}+\frac{1}{8\ln\left(2\right)}=\frac{5}{8\ln\left(2\right)}\cong0.90168 \end{gathered}[/tex]The value of the integral is 0.90168 and represents the area of the region.
