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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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_{1}^{2} \frac{1 + ln ( x )}{x} d x

\int_{4}^{3} sin (t) d t

\int_{0}^{1} 3 (1 - y)^{2} d y

\int_{0}^{1} e^{- s x} d x

\int_{0}^{t} \frac{1}{t} d t

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