What is the integral from 0 to 1 of tsqrt(1+4t^2) ?

The integral from 0 to 1 of tsqrt(1+4t^2) is (5sqrt(5)-1)/(12)

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integral from 0 to 1 of tsqrt(1+4t^2)

\int_{0}^{1} t 1 + 4 t^{2} d t

\frac{5 5 - 1}{12}

Decimal Notation

0.84836 \dots

Solution steps

\int_{0}^{1} t 1 + 4 t^{2} d t

Apply u-substitution

: \int_{1}^{5} \frac{u}{8} d u

= \int_{1}^{5} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{10 5}{3} - \frac{2}{3})

Simplify

= \frac{5 5 - 1}{12}

x y^{^{'}} = y + e^{\frac{y}{x}}

x \to 2 lim (- 2)

\frac{d x}{d t} = x sin (t), x (0) = 1

\int 6 tan (x) sec^{3} (x) d x

f (x) = 8 cos (x)

What is the integral from 0 to 1 of tsqrt(1+4t^2) ?
The integral from 0 to 1 of tsqrt(1+4t^2) is (5sqrt(5)-1)/(12)