What is the integral from 0 to 2pi of sqrt(36+9/4 t) ?

The integral from 0 to 2pi of sqrt(36+9/4 t) is (6pisqrt(16+2pi)-(2pi+16)^{3/2}+48sqrt(2pi+16))/2-64

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integral from 0 to 2pi of sqrt(36+9/4 t)

\int_{0}^{2 π} 36 + \frac{9}{4} t d t

\frac{6 π 16 + 2 π - ( 2 π + 16 ) ^{\frac{3}{2}} + 48 2 π + 16}{2} - 64

Solution steps

\int_{0}^{2 π} 36 + \frac{9}{4} t d t

Apply Integration By Parts

: [\frac{1}{2} (3 t 16 + t - 2 \cdot \int \frac{3 t}{4 t + 16} d t)]_{0}^{2 π}

= [\frac{1}{2} (3 t 16 + t - 2 \cdot \int \frac{3 t}{4 t + 16} d t)]_{0}^{2 π}

\int \frac{3 t}{4 t + 16} d t = \frac{1}{2} (t + 16)^{\frac{3}{2}} - 24 t + 16

= [\frac{1}{2} (3 t 16 + t - 2 (\frac{1}{2} (t + 16)^{\frac{3}{2}} - 24 t + 16))]_{0}^{2 π}

Simplify

[\frac{1}{2} (3 t 16 + t - 2 (\frac{1}{2} (t + 16)^{\frac{3}{2}} - 24 t + 16))]_{0}^{2 π} : [\frac{1}{2} (3 t 16 + t - (t + 16)^{\frac{3}{2}} + 48 t + 16)]_{0}^{2 π}

= [\frac{1}{2} (3 t 16 + t - (t + 16)^{\frac{3}{2}} + 48 t + 16)]_{0}^{2 π}

Compute the boundaries:

\frac{6 π 16 + 2 π - ( 2 π + 16 ) ^{\frac{3}{2}} + 48 2 π + 16}{2} - 64

= \frac{6 π 16 + 2 π - ( 2 π + 16 ) ^{\frac{3}{2}} + 48 2 π + 16}{2} - 64

\int \frac{2 x + 6}{x ^{2} + 2 x + 5} d x

(1 + 2 x)^{10}

\int \frac{1}{( π \cdot x ) ^{10}} d x

\frac{d}{d x} (4 + x x)

\frac{x ^{2}}{1 - x ^{2}}

What is the integral from 0 to 2pi of sqrt(36+9/4 t) ?
The integral from 0 to 2pi of sqrt(36+9/4 t) is (6pisqrt(16+2pi)-(2pi+16)^{3/2}+48sqrt(2pi+16))/2-64