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

The integral from 0 to 1 of sqrt(1+(2y)^2) is 1/2 (sqrt(5)+1/2 ln(2+sqrt(5)))

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integral from 0 to 1 of sqrt(1+(2y)^2)

\int_{0}^{1} 1 + (2 y)^{2} d y

\frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

Decimal

1.47894 \dots

Solution steps

\int_{0}^{1} 1 + (2 y)^{2} d y

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2)} \frac{1}{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{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

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

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Compute the boundaries:

5

= \frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

\int_{- 5}^{3} x^{2} + x - 12 d x

\int_{- 1}^{1} \frac{1}{( x ^{2} + 1 )} d x

\int_{0}^{5.4} 378 - 17 t^{2} d t

\int_{0}^{4} (x^{2} - 4 x + 4)^{2} d x

\int_{0}^{\frac{1}{2}} 5 cos (π) x d x

What is the integral from 0 to 1 of sqrt(1+(2y)^2) ?
The integral from 0 to 1 of sqrt(1+(2y)^2) is 1/2 (sqrt(5)+1/2 ln(2+sqrt(5)))