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

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

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

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

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

Decimal

9.14912 \dots

Solution steps

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

Take the constant out:

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

= 2 \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}

= 2 [\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}

= 2 [\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}

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

Compute the boundaries:

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

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

Simplify

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

u^{3}

\frac{d}{d x} (12 + 4 x - x^{2})

\frac{d}{d x} (1 + 2 x - x^{3})

\int (\frac{1}{( 1 + x ) ^{2}}) d x

\int_{1}^{\infty} x e^{- 3 x} d x

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