What is the integral from-1 to 2 of x^2+e^{2x} ?

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

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integral from-1 to 2 of x^2+e^{2x}

\int_{- 1}^{2} x^{2} + e^{2 x} d x

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

Decimal

30.23140 \dots

Solution steps

\int_{- 1}^{2} x^{2} + e^{2 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}^{2} x^{2} d x + \int_{- 1}^{2} e^{2 x} d x

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

\int_{- 1}^{2} e^{2 x} d x = \frac{e ^{6} - 1}{2 e ^{2}}

= 3 + \frac{e ^{6} - 1}{2 e ^{2}}

\int_{- 2}^{3} (2 x^{4} + 5 x^{3} - x + 5) d x

\int_{- 2}^{0} x 2 x^{2} + 1 d x

\int_{0}^{1} (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

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

\int_{1}^{5} \frac{1}{96} x y d y

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