What is the integral of (-3x+4)e-x ?

The integral of (-3x+4)e-x is e(-(3x^2)/2+4x)-(x^2)/2+C

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integral of (-3x+4)e-x

\int (- 3 x + 4) e - x d x

e (- \frac{3 x ^{2}}{2} + 4 x) - \frac{x ^{2}}{2} + C

Solution steps

\int (- 3 x + 4) e - x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int (- 3 x + 4) e d x - \int x d x

\int (- 3 x + 4) e d x = e (- \frac{3 x ^{2}}{2} + 4 x)

\int x d x = \frac{x ^{2}}{2}

= e (- \frac{3 x ^{2}}{2} + 4 x) - \frac{x ^{2}}{2}

Add a constant to the solution

= e (- \frac{3 x ^{2}}{2} + 4 x) - \frac{x ^{2}}{2} + C

\int_{0}^{\frac{π}{2}} 7 cos^{2} (x) d x

y = x^{2}, y = - x^{2} + 18 x

f (x) = c x^{- 7}

\int_{0}^{\frac{π}{2}} sin^{3} (x) d x

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What is the integral of (-3x+4)e-x ?
The integral of (-3x+4)e-x is e(-(3x^2)/2+4x)-(x^2)/2+C