What is the integral from 0 to 0.4 of sqrt(1+x^2) ?

The integral from 0 to 0.4 of sqrt(1+x^2) is 0.41042…

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integral from 0 to 0.4 of sqrt(1+x^2)

\int_{0}^{0.4} 1 + x^{2} d x

0.41042 \dots

Solution steps

\int_{0}^{0.4} 1 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{0.38050 \dots} sec^{3} (u) d u

= \int_{0}^{0.38050 \dots} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{0.38050 \dots} + \frac{1}{2} \cdot \int_{0}^{0.38050 \dots} sec (u) d u

\int_{0}^{0.38050 \dots} sec (u) d u = 0.39003 \dots

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{0.38050 \dots} + \frac{1}{2} \cdot 0.39003 \dots

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{0.38050 \dots} + \frac{1}{2} \cdot 0.39003 \dots : [0.5 sec^{2} (u) sin (u)]_{0}^{0.38050 \dots} + 0.19501 \dots

= [0.5 sec^{2} (u) sin (u)]_{0}^{0.38050 \dots} + 0.19501 \dots

Compute the boundaries:

0.21540 \dots

= 0.21540 \dots + 0.19501 \dots

Simplify

= 0.41042 \dots

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What is the integral from 0 to 0.4 of sqrt(1+x^2) ?
The integral from 0 to 0.4 of sqrt(1+x^2) is 0.41042…