What is the integral from 1 to e of 4t^2ln(t) ?

The integral from 1 to e of 4t^2ln(t) is (4(2e^3+1))/9

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integral from 1 to e of 4t^2ln(t)

\int_{1}^{e} 4 t^{2} ln (t) d t

\frac{4 ( 2 e ^{3} + 1 )}{9}

Decimal

18.29825 \dots

Solution steps

\int_{1}^{e} 4 t^{2} ln (t) d t

Take the constant out:

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

= 4 \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}

= 4 [\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}

= 4 [\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}

= 4 [\frac{1}{3} (t^{3} ln (t) - \frac{t ^{3}}{3})]_{1}^{e}

Compute the boundaries:

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

= 4 (\frac{2 e ^{3}}{9} + \frac{1}{9})

Simplify

= \frac{4 ( 2 e ^{3} + 1 )}{9}

x \to 0 lim (tan^{x} (5 x))

4 x^{2} - 5 x + 1

\frac{\partial}{\partial x} (\frac{x y}{y})

\frac{\partial}{\partial x} (x + 2 y)

\int_{1}^{e} ln (22 x) d x

What is the integral from 1 to e of 4t^2ln(t) ?
The integral from 1 to e of 4t^2ln(t) is (4(2e^3+1))/9