Answer
$$\int^1_0(5x-5^x)dx=\frac{5}{2}-\frac{4}{\ln5}$$
Work Step by Step
$$A=\int^1_0(5x-5^x)dx$$ According to Table 1, we have $$\int x^ndx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int b^xdx=\frac{b^x}{\ln|b|}+C$$
Therefore, $$A=(\frac{5x^2}{2}-\frac{5^x}{\ln5})\Bigg]^1_0$$ $$A=(\frac{5\times1^2}{2}-\frac{5^1}{\ln5})-(\frac{5\times0^2}{2}-\frac{5^0}{\ln5})$$ $$A=(\frac{5}{2}-\frac{5}{\ln5})-(0-\frac{1}{\ln5})$$ $$A=\frac{5}{2}-\frac{5}{\ln5}+\frac{1}{\ln5}$$ $$A=\frac{5}{2}-\frac{4}{\ln5}$$
