What is the integral from 2 to 4 of e^{tx} x/6 ?

The integral from 2 to 4 of e^{tx} x/6 is (t(4e^{4t}-2e^{2t})-e^{4t}+e^{2t})/(6t^2)

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integral from 2 to 4 of e^{tx} x/6

\int_{2}^{4} e^{t x} \frac{x}{6} d x

\frac{t ( 4 e ^{4 t} - 2 e ^{2 t} ) - e ^{4 t} + e ^{2 t}}{6 t ^{2}}

Solution steps

\int_{2}^{4} e^{t x} \frac{x}{6} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int_{2}^{4} x e^{t x} d x

Apply Integration By Parts

: [e^{t x} \frac{1}{t} x - \int e^{t x} \frac{1}{t} d x]_{2}^{4}

= \frac{1}{6} [e^{t x} \frac{1}{t} x - \int e^{t x} \frac{1}{t} d x]_{2}^{4}

\int e^{t x} \frac{1}{t} d x = e^{t x} \frac{1}{t ^{2}}

= \frac{1}{6} [e^{t x} \frac{1}{t} x - e^{t x} \frac{1}{t ^{2}}]_{2}^{4}

Compute the boundaries:

e^{4 t} \frac{4}{t} - e^{4 t} \frac{1}{t ^{2}} - e^{2 t} \frac{2}{t} + e^{2 t} \frac{1}{t ^{2}}

= \frac{1}{6} (e^{4 t} \frac{4}{t} - e^{4 t} \frac{1}{t ^{2}} - e^{2 t} \frac{2}{t} + e^{2 t} \frac{1}{t ^{2}})

Simplify

= \frac{t ( 4 e ^{4 t} - 2 e ^{2 t} ) - e ^{4 t} + e ^{2 t}}{6 t ^{2}}

\frac{d}{d x} (x^{2} + y^{2} - 2 x y)

\int \frac{1}{t ^{2}} d t

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\frac{\partial}{\partial x} (10 cos (x) y^{2} z)

3 x^{2} e^{x} + e^{x} (x^{3} + 6)

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