What is the integral from 1 to 4 of sqrt(t)ln(t/2) ?

The integral from 1 to 4 of sqrt(t)ln(t/2) is 6ln(2)-28/9

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integral from 1 to 4 of sqrt(t)ln(t/2)

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

6 ln (2) - \frac{28}{9}

Decimal Notation

1.04777 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{2}{3} t^{\frac{3}{2}} ln (\frac{t}{2}) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{4}

= [\frac{2}{3} t^{\frac{3}{2}} ln (\frac{t}{2}) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{4}

\int \frac{2}{3} t^{\frac{1}{2}} d t = \frac{4}{9} t^{\frac{3}{2}}

= [\frac{2}{3} t^{\frac{3}{2}} ln (\frac{t}{2}) - \frac{4}{9} t^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

6 ln (2) - \frac{28}{9}

= 6 ln (2) - \frac{28}{9}

\int \frac{4 x}{1 - 16 x ^{4}} d x

y = \frac{x}{a} + \frac{a}{x}, at x = \frac{a}{2}

x \to 5 lim ([x])

f (x) = x^{2} + 8, (- 3, 17)

x \to - 2 lim (4 x)

What is the integral from 1 to 4 of sqrt(t)ln(t/2) ?
The integral from 1 to 4 of sqrt(t)ln(t/2) is 6ln(2)-28/9