What is the integral from 0 to 1 of 6tsqrt(1+t^2) ?

The integral from 0 to 1 of 6tsqrt(1+t^2) is 2(2sqrt(2)-1)

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integral from 0 to 1 of 6tsqrt(1+t^2)

\int_{0}^{1} 6 t 1 + t^{2} d t

2 (22 - 1)

Decimal

3.65685 \dots

Solution steps

\int_{0}^{1} 6 t 1 + t^{2} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{1} t 1 + t^{2} d t

Apply u-substitution

: \int_{1}^{2} u^{2} d u

= 6 \cdot \int_{1}^{2} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{2}

= 6 [\frac{u ^{3}}{3}]_{1}^{2}

Compute the boundaries:

\frac{2 2}{3} - \frac{1}{3}

= 6 (\frac{2 2}{3} - \frac{1}{3})

Simplify

= 2 (22 - 1)

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What is the integral from 0 to 1 of 6tsqrt(1+t^2) ?
The integral from 0 to 1 of 6tsqrt(1+t^2) is 2(2sqrt(2)-1)