What is the integral from 1 to e of 5t^2ln(t) ?

The integral from 1 to e of 5t^2ln(t) is (5(2e^3+1))/9

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integral from 1 to e of 5t^2ln(t)

\int_{1}^{e} 5 t^{2} ln (t) d t

\frac{5 ( 2 e ^{3} + 1 )}{9}

Decimal Notation

22.87281 \dots

Solution steps

\int_{1}^{e} 5 t^{2} ln (t) d t

Take the constant out:

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

= 5 \cdot \int_{1}^{e} t^{2} ln (t) d t

Apply Integration By Parts

: [\frac{1}{3} (t^{3} ln (t) - 3 \cdot \int \frac{t ^{2}}{3} d t)]_{1}^{e}

= 5 [\frac{1}{3} (t^{3} ln (t) - 3 \cdot \int \frac{t ^{2}}{3} d t)]_{1}^{e}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= 5 [\frac{1}{3} (t^{3} ln (t) - 3 \cdot \frac{t ^{3}}{9})]_{1}^{e}

3 \cdot \frac{t ^{3}}{9} = \frac{t ^{3}}{3}

= 5 [\frac{1}{3} (t^{3} ln (t) - \frac{t ^{3}}{3})]_{1}^{e}

Compute the boundaries:

\frac{2 e ^{3}}{9} + \frac{1}{9}

= 5 (\frac{2 e ^{3}}{9} + \frac{1}{9})

Simplify

= \frac{5 ( 2 e ^{3} + 1 )}{9}

\frac{d x}{d t} = x - x^{2}, x (0) = 2

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} - 6 x = 4 e^{- 2 t}

\int - 36 cos (x) d x

\int 1 + y d y

x \to 0 lim (\frac{3 x ^{2}}{x})

What is the integral from 1 to e of 5t^2ln(t) ?
The integral from 1 to e of 5t^2ln(t) is (5(2e^3+1))/9