What is the integral from 0 to 1 of xe^{10x} ?

The integral from 0 to 1 of xe^{10x} is (9e^{10}+1)/(100)

integral from 0 to 1 of xe^{10x}

\int_{0}^{1} x e^{10 x} d x

\frac{9 e ^{10} + 1}{100}

Decimal

1982.39192 \dots

Solution steps

\int_{0}^{1} x e^{10 x} d x

Apply u-substitution

= \int_{0}^{10} \frac{e ^{u} u}{100} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{100} \cdot \int_{0}^{10} e^{u} u d u

Apply Integration By Parts

= \frac{1}{100} [e^{u} u - \int e^{u} d u]_{0}^{10}

\int e^{u} d u = e^{u}

= \frac{1}{100} [e^{u} u - e^{u}]_{0}^{10}

Compute the boundaries:

9 e^{10} + 1

= \frac{1}{100} (9 e^{10} + 1)

Simplify

= \frac{9 e ^{10} + 1}{100}

y^{^{''}} - 4 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

d er i v a t i v e f (x) = x^{2} [0.2]

s l o p e (- 10, 8), (- 15, - 7)

\int \frac{x}{2 - x} d x

d er i v a t i v e \frac{2 x ^{2}}{( 3 - 2 x ) ^{4}}

What is the integral from 0 to 1 of xe^{10x} ?
The integral from 0 to 1 of xe^{10x} is (9e^{10}+1)/(100)