What is the integral from 1 to e of (4e^x+4sin(x)) ?

The integral from 1 to e of (4e^x+4sin(x)) is 4(e^e-e)+4(-cos(e)+cos(1))

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integral from 1 to e of (4e^x+4sin(x))

\int_{1}^{e} (4 e^{x} + 4 sin (x)) d x

4 (e^{e} - e) + 4 (- cos (e) + cos (1))

Decimal

55.55206 \dots

Solution steps

\int_{1}^{e} 4 e^{x} + 4 sin (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_{1}^{e} 4 e^{x} d x + \int_{1}^{e} 4 sin (x) d x

\int_{1}^{e} 4 e^{x} d x = 4 (e^{e} - e)

\int_{1}^{e} 4 sin (x) d x = 4 (- cos (e) + cos (1))

= 4 (e^{e} - e) + 4 (- cos (e) + cos (1))

\int_{- \frac{π}{4}}^{\frac{π}{4}} csc^{2} (x) d x

\int_{1}^{3} x^{2} - 2 x - 4 d x

\int_{0}^{4} (3 x^{2} + 4 x + 7) d x

\int_{5}^{8} x^{2} d x

\int_{0}^{6} \frac{x}{9 + 2 x ^{2}} d x

What is the integral from 1 to e of (4e^x+4sin(x)) ?
The integral from 1 to e of (4e^x+4sin(x)) is 4(e^e-e)+4(-cos(e)+cos(1))