Calculate $\int^3_0 \frac{\ln2}{2} \left(2^{x-3}-2^{3-x}\right)$

Question

$$\begin{align} Definite\,integral & = \int^3_0 \frac{\ln2}{2} \left(2^{x-3}-2^{3-x}\right)\\ & = \frac12\int^3_0 \ln 2 \left(2^{x-3}-2^{3-x}\right)\\ & = \frac12\int^3_0 \ln 2 \cdot 2^{x-3}-\ln2\cdot2^{3-x}\\ & = \frac12\left[2^{x-3}-2^{3-x}\right]^3_0\\ & = 0.5\left(0-\left(-\frac{63}{8}\right)\right) = 3\frac{15}{16} \end{align}$$

However, according to my book and Wolfram Alpha, this should be $-3 \frac{1}{16}$.

Does someone spot my mistake? Thanks!

Answer 1

$$\int \ln2 \cdot 2^{3-x}=\color{red}-2^{3-x} $$

Therefore : $$\int^3_0 \frac{\ln2}{2} \left(2^{x-3}-2^{3-x}\right)$$ \begin{align} &= \frac12\int^3_0 \ln 2 \cdot 2^{x-3}-\ln2\cdot2^{3-x}\\ & = \frac12\left[2^{x-3}\color{red}+2^{3-x}\right]^3_0\\ & = \frac12\left(2-\left(\frac{65}{8}\right)\right) = -3\frac{1}{16} \end{align}

as provided the answer in your textbook.