What is the integral from 0 to 20 of sqrt(1+4x^2) ?

The integral from 0 to 20 of sqrt(1+4x^2) is 1/2 (20sqrt(1601)+1/2 ln(40+sqrt(1601)))

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integral from 0 to 20 of sqrt(1+4x^2)

\int_{0}^{20} 1 + 4 x^{2} d x

\frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

Decimal

401.22052 \dots

Solution steps

\int_{0}^{20} 1 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (40)} \frac{1}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (40)} \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 (40)} 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 (40)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (40)} sec (u) d u)

\int_{0}^{a r c t a n (40)} sec (u) d u = ln (40 + 1601)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Compute the boundaries:

201601

= \frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

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What is the integral from 0 to 20 of sqrt(1+4x^2) ?
The integral from 0 to 20 of sqrt(1+4x^2) is 1/2 (20sqrt(1601)+1/2 ln(40+sqrt(1601)))