How do you evaluate the integral $\int 3 e^{x} - 5 e^{2 x} d x$ $int3e^(x)-5e^(2x) dx$ ?

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Answer 1 · 2014-09-25T20:12:47

$\int 3 e^{x} - 5 e^{2 x} d x = 3 e^{x} - \frac{5}{2} e^{2 x} + C$ $int 3e^x-5e^{2x}dx=3e^x-5/2e^{2x}+C$

Let us look at some details.

Remember:

$\int e^{x} d x = e^{x} + C$ $int e^x dx=e^x+C$

$\int e^{k x} d x = \frac{e^{k x}}{k} + C$ $int e^{kx} dx=e^{kx}/k+C$

Now, let us work on the integral.

$\int 3 e^{x} - 5 e^{2 x} d x$ $int 3e^x-5e^{2x}dx$

by applying integral on each term,

$= \int 3 e^{x} d x - \int 5 e^{2 x} d x$ $=int 3e^x dx -int 5e^{2x}dx$

by pulling constants out of the integrals

$= 3 \int e^{x} d x - 5 \int e^{2 x} d x$ $=3int e^x dx-5int e^{2x}dx$

by applying the formulas above,

$= 3 e^{x} - 5 \frac{e^{2 x}}{2} + C$ $=3e^x-5e^{2x}/2+C$

How do you evaluate the integral ∫3ex−5e2xdxint3e^(x)-5e^(2x) dx?