The integral of
e−x is
−e−x+C.
Solution. We want to determine the integral of
e−x, i.e.:
∫ex1dx=∫e−xdx.
We will use the substitution method: let
u=−x, then
du=−dx. Then we get the following integral:
∫e−xdx=−∫eudu=−eu+C=−e−x+C.
Therefore, the integral of
e−x is
−e−x+C.