Объяснение:

[tex]S=\int\limits^2_{-3} {((\frac{1}{2})^x-0) } \, dx =\int\limits^2_{-3} {(\frac{1}{2})^x } \, dx=\int\limits {(2)^{-x }} \, dx.\\\int\limits {(2)^{-x }} \, dx=|\left {{-x=u\ \ \ x=-u} \atop {dx=-du}} \right. |=-\int 2^u du=-\frac{2^u}{ln2}+C=-\frac{2^{-x}}{ln2} +C=-\frac{1}{ln2*2^x}+C\\ S=(-\frac{1}{ln2*2^x})\ |_{-3}^2=-(\frac{1}{ln2*2^2}-\frac{1}{ln2*2^{-3}} )}=-(\frac{1}{ln2*4}-\frac{8}{ln2})=\\=\frac{32}{4*ln2}-\frac{1}{4*ln2}=\frac{31}{4*ln2}.[/tex]