Respuesta :

You can start with the indefinite integral

... ∫e^(ax)·dx = (1/a)e^(ax) +c


Then your definite integral is

[tex]\int\limits_{-3}^{10}{e^{-0.025x}}\,dx=\dfrac{-1}{.025}\left(e^{-0.025\cdot 10}-e^{-0.025\cdot(-3)}\right)\\\\=40\left(e^{.075}-e^{-0.25}\right)\\\\ \approx11.9633[/tex]

1rstar
Hey there !

Check the attachment.
Hope it helps you :)
Ver imagen 1rstar