What is the integral from 1 to 2 of tsqrt(1/2 t^2+8) ?

The integral from 1 to 2 of tsqrt(1/2 t^2+8) is (20sqrt(10))/3-(17sqrt(17))/(3sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of tsqrt(1/2 t^2+8)

\int_{1}^{2} t \frac{1}{2} t^{2} + 8 d t

\frac{20 10}{3} - \frac{17 17}{3 2}

Decimal

4.56082 \dots

Solution steps

\int_{1}^{2} t \frac{1}{2} t^{2} + 8 d t

Apply Integration By Parts

: [\frac{1}{2 2} t^{2} t^{2} + 16 - \int \frac{t ^{3}}{2 2 t ^{2} + 16} d t]_{1}^{2}

= [\frac{1}{2 2} t^{2} t^{2} + 16 - \int \frac{t ^{3}}{2 2 t ^{2} + 16} d t]_{1}^{2}

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

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

Simplify

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

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

Compute the boundaries:

\frac{20 10}{3} - \frac{17 17}{3 2}

= \frac{20 10}{3} - \frac{17 17}{3 2}

\int_{a}^{b} 3 x^{2} d x

\int_{0}^{1} (e^{x} + 1) e^{x} d x

\int_{0}^{4} 9 x^{2} d x

\int_{0}^{y} \frac{6 ( y - x )}{8} d x

\int_{0}^{2} (2.5 - x^{2}) d x

What is the integral from 1 to 2 of tsqrt(1/2 t^2+8) ?
The integral from 1 to 2 of tsqrt(1/2 t^2+8) is (20sqrt(10))/3-(17sqrt(17))/(3sqrt(2))