Question

Evaluate the integral \[\int\limits_0^1 {\sqrt {{x^2} + 1} } \,\,dx\].

Short answer

The Evaluation of Equation gives us ,\[\int\limits_0^1 {\sqrt {{x^2} + 1} } \,\,dx = 0.8012 + c\]

Step-by-step solution

Find the integral.

\[\int\limits_0^1 {\sqrt {{x^2} + 1} } \,\,dx\]

Let,

\[\begin{array}{l}x = tan\theta \\dx = se{c^2}\theta \,\,d\theta \\\int\limits_0^1 {\sqrt {{x^2} + 1} } \,\,dx = \int\limits_0^1 {\sqrt {ta{n^2}\theta + 1} } se{c^2}\theta \,\,d\theta \end{array}\]

Using \[ta{n^2}\theta + 1 = se{c^2}\theta \]

\[\begin{array}{l} = \int\limits_0^1 {\sqrt {se{c^2}\theta } } se{c^2}\theta \,d\theta \\ = \int\limits_0^1 {sec\theta \,se{c^2}\theta \,d\theta } \\ = \int\limits_0^1 {se{c^3}\theta \,d\theta } \\ = \left[ {\frac{1}{2}sec\theta \,tan\theta + \frac{1}{2}ln\left| {sec\theta + tan\theta } \right|} \right]_0^1 + c\end{array}\]

Changing \[\theta \]to \[x\], We know \[sec\theta = \sqrt {{x^2} + 1} ,tan\theta = x\]

\[\begin{array}{l} = \left[ {\frac{1}{2}\sqrt {{x^2} + 1} \,x + \frac{1}{2}ln\left| {\sqrt {{x^2} + 1} + x} \right|} \right]_0^1 + c\\ = \left[ {\frac{1}{2}x\sqrt {{x^2} + 1} + \frac{1}{2}ln\left( {\sqrt {{x^2} + 1} + x} \right)} \right]_0^1 + c\end{array}\]

To put an upper and lower limit.

\[\begin{array}{l}\int\limits_0^1 {\sqrt {{x^2} + 1} } dx = \left[ {\frac{1}{2}x\sqrt {{x^2} + 1} + \frac{1}{2}ln\left( {\sqrt {{x^2} + 1} + x} \right)} \right]_0^1 + c\\ = \frac{1}{2} \times 1\sqrt {{1^2} + 1} + \frac{1}{2}ln\left( {\sqrt {{1^2} + 1} + 1} \right) - \left( {\frac{1}{2} \times 0\sqrt {{0^2} + 1} + \frac{1}{2}ln\left( {\sqrt {{0^2} + 1} + 1} \right)} \right)\\ = \frac{1}{2}\sqrt 2 + \frac{1}{2}ln\left( {\sqrt 2 + 1} \right) - \left( {0 + \frac{1}{2}ln\left( {\sqrt 1 + 1} \right)} \right)\\ = \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times 0.8813 - \frac{1}{2} \times 0.6931\\ = \frac{1}{{\sqrt 2 }} + \frac{{0.8813}}{2} - \frac{{0.6931}}{2}\\ = 0.7071 + 0.44005 - 0.34655\\ = 0.8012 + c\end{array}\]

Hence, \[\int\limits_0^1 {\sqrt {{x^2} + 1} } \,dx = 0.8012 + c\]