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

The integral from 1 to 2e of 4ln(x) is 4(2eln(2e)-2e+1)

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integral from 1 to 2e of 4ln(x)

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

4 (2 e ln (2 e) - 2 e + 1)

Decimal

19.07335 \dots

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{1}^{2 e} ln (x) d x

Apply Integration By Parts

= 4 [x ln (x) - \int 1 d x]_{1}^{2 e}

\int 1 d x = x

= 4 [x ln (x) - x]_{1}^{2 e}

Compute the boundaries:

2 e ln (2 e) - 2 e + 1

= 4 (2 e ln (2 e) - 2 e + 1)

\int_{- 1}^{1} - x^{2} + 2 - x^{2} d x

\int_{0}^{1} \frac{x ^{3}}{2 ( 1 + x ^{4} )} d x

\int_{- 4}^{4} x^{3} - 16 x d x

\int_{1}^{2} x^{3} - x^{2} d x

\int_{- 2}^{1} x + 2 d x

What is the integral from 1 to 2e of 4ln(x) ?
The integral from 1 to 2e of 4ln(x) is 4(2eln(2e)-2e+1)