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

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

integral from 1 to e of 3t^2ln(t)

\int_{1}^{e} 3 t^{2} ln (t) d t

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

Decimal

13.72369 \dots

Solution steps

\int_{1}^{e} 3 t^{2} ln (t) d t

Take the constant out:

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

= 3 \cdot \int_{1}^{e} t^{2} ln (t) d t

Apply Integration By Parts

= 3 [\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}

= 3 [\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}

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

Compute the boundaries:

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

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

Simplify

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

\int_{- 2}^{3} (8 x - 4) d x

\int_{- 6 π}^{6 π} cos^{2} (x) d x

\int_{1}^{e} \frac{t ^{6} + 1}{t} d t

\int_{0}^{π} (2 e^{x} + 3 sin (x)) d x

\int_{0}^{l n (2)} (e^{x} - e^{- x}) d x

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