What is the integral from 0 to 1 of (x^2+4)e^{3x} ?

The integral from 0 to 1 of (x^2+4)e^{3x} is (41e^3-38)/(27)

Popular >

integral from 0 to 1 of (x^2+4)e^{3x}

\int_{0}^{1} (x^{2} + 4) e^{3 x} d x

\frac{41 e ^{3} - 38}{27}

Decimal Notation

29.09285 \dots

Solution steps

\int_{0}^{1} (x^{2} + 4) e^{3 x} d x

Expand

(x^{2} + 4) e^{3 x} : e^{3 x} x^{2} + 4 e^{3 x}

= \int_{0}^{1} e^{3 x} x^{2} + 4 e^{3 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_{0}^{1} e^{3 x} x^{2} d x + \int_{0}^{1} 4 e^{3 x} d x

\int_{0}^{1} e^{3 x} x^{2} d x = \frac{5 e ^{3} - 2}{27}

\int_{0}^{1} 4 e^{3 x} d x = \frac{4 ( e ^{3} - 1 )}{3}

= \frac{5 e ^{3} - 2}{27} + \frac{4 ( e ^{3} - 1 )}{3}

Simplify

\frac{5 e ^{3} - 2}{27} + \frac{4 ( e ^{3} - 1 )}{3} : \frac{41 e ^{3} - 38}{27}

= \frac{41 e ^{3} - 38}{27}

\int_{0}^{4} x x^{2} + 6 d x

\int_{0}^{1} x^{2} (2 + 2 x^{3})^{5} d x

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

\int_{0}^{3} x^{2} e^{x^{3}} d x

\int_{0}^{π} - 2 cos^{2} (x) sin (x) d x

What is the integral from 0 to 1 of (x^2+4)e^{3x} ?
The integral from 0 to 1 of (x^2+4)e^{3x} is (41e^3-38)/(27)