What is the integral from 1 to 2 of (e^t+4t) ?

The integral from 1 to 2 of (e^t+4t) is e^2-e+6

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integral from 1 to 2 of (e^t+4t)

\int_{1}^{2} (e^{t} + 4 t) d t

e^{2} - e + 6

Decimal

10.67077 \dots

Solution steps

\int_{1}^{2} e^{t} + 4 t d t

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}^{2} e^{t} d t + \int_{1}^{2} 4 t d t

\int_{1}^{2} e^{t} d t = e^{2} - e

\int_{1}^{2} 4 t d t = 6

= e^{2} - e + 6

\int_{0}^{1} x^{15} + 1 5^{x} d x

\int_{- 1}^{1} 1 - t d t

\int_{0}^{1 - z - y} z d z

\int_{1}^{3} - \frac{1}{t ^{3}} - 3 d t

\int_{0}^{1} \frac{11}{1 + p ^{2}} d p

What is the integral from 1 to 2 of (e^t+4t) ?
The integral from 1 to 2 of (e^t+4t) is e^2-e+6