What is the integral from-6 to 6 of 3t+4e^{2t}-e^t ?

The integral from-6 to 6 of 3t+4e^{2t}-e^t is 2e^{12}-e^6+1/(e^6)-2/(e^{12)}

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integral from-6 to 6 of 3t+4e^{2t}-e^t

\int_{- 6}^{6} 3 t + 4 e^{2 t} - e^{t} d t

2 e^{12} - e^{6} + \frac{1}{e ^{6}} - \frac{2}{e ^{12}}

Decimal Notation

325106.15651 \dots

Solution steps

\int_{- 6}^{6} 3 t + 4 e^{2 t} - e^{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_{- 6}^{6} 3 t d t + \int_{- 6}^{6} 4 e^{2 t} d t - \int_{- 6}^{6} e^{t} d t

\int_{- 6}^{6} 3 t d t = 0

\int_{- 6}^{6} 4 e^{2 t} d t = 2 (e^{12} - \frac{1}{e ^{12}})

\int_{- 6}^{6} e^{t} d t = e^{6} - \frac{1}{e ^{6}}

= 0 + 2 (e^{12} - \frac{1}{e ^{12}}) - (e^{6} - \frac{1}{e ^{6}})

Simplify

0 + 2 (e^{12} - \frac{1}{e ^{12}}) - (e^{6} - \frac{1}{e ^{6}}) : 2 e^{12} - e^{6} + \frac{1}{e ^{6}} - \frac{2}{e ^{12}}

= 2 e^{12} - e^{6} + \frac{1}{e ^{6}} - \frac{2}{e ^{12}}

\frac{\partial}{\partial y} (x yz + sin (2 x y))

5 cos (π x), 12 x^{3} - 3, 0.5, - 0.5

\frac{1}{1 + x}, 1

f (x) = x^{3} - 6 x, g (x) = - 2 x

36 t^{2}

What is the integral from-6 to 6 of 3t+4e^{2t}-e^t ?
The integral from-6 to 6 of 3t+4e^{2t}-e^t is 2e^{12}-e^6+1/(e^6)-2/(e^{12)}