What is the integral from 0 to 1 of sqrt(7+4t^2) ?

The integral from 0 to 1 of sqrt(7+4t^2) is 1/4 (2sqrt(11)+7ln(2/(sqrt(7))+sqrt(11/7)))

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integral from 0 to 1 of sqrt(7+4t^2)

\int_{0}^{1} 7 + 4 t^{2} d t

\frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

Decimal

2.87960 \dots

Solution steps

\int_{0}^{1} 7 + 4 t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{2}{7})} \frac{7}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{2}{7})} \frac{7}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{7}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{7})} sec^{3} (u) d u

Apply Integral Reduction

= \frac{7}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{7})} sec (u) d u)

\int_{0}^{a r c t a n (\frac{2}{7})} sec (u) d u = ln (\frac{2}{7} + \frac{11}{7})

= \frac{7}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7}))

Simplify

\frac{7}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7})) : \frac{1}{4} (14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7}))

= \frac{1}{4} (14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7}))

Compute the boundaries:

\frac{11}{7}

= \frac{1}{4} (14 \cdot \frac{11}{7} + 7 ln (\frac{2}{7} + \frac{11}{7}))

Simplify

= \frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

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What is the integral from 0 to 1 of sqrt(7+4t^2) ?
The integral from 0 to 1 of sqrt(7+4t^2) is 1/4 (2sqrt(11)+7ln(2/(sqrt(7))+sqrt(11/7)))