Question

Find the length of the curve y = 3 + x 2 from (0 3) to (2 4)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The length of the given curve is 9.156 units.

Step-by-step explanation:

As per the question, we have given the equation

$y=3+x^2$

and coordinates of the curve are (0, 3) and (2, 4).

First, we will differentiate the equation of the curve with respect to x, then we will use the formula of the length of the curve taking the integration limits from 0 to 3.

Step 1

$y=3+x^2$

$\[\frac{{dy}}{{dx}} = 2x\]$

Step 2

As we know the formula for the length of the curve is

The required length of curve = $\[\int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \]$

Therefore,

Required length = $\[\int\limits_0^2 {\sqrt {1 + {{\left( {2x} \right)}^2}} dx} \]$

As we know that

$\[\int {\sqrt {{x^2} + {a^2}} } = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right|\]$

Therefore,

$\[\int_0^2 {\sqrt {{{(2x)}^2} + {1^2}} } dx = \left[ {\frac{{2x}}{2}\sqrt {{{(2x)}^2} + {1^2}} + \frac{{{1^2}}}{2}\log \left| {2x + \sqrt {{{(2x)}^2} + {1^2}} } \right|} \right]_0^2\]$

$\[\int_0^2 {\sqrt {{{(2x)}^2} + {1^2}} } dx = \frac{{2 \times 2}}{2}\sqrt {{{(2 \times 2)}^2} + {1^2}} + \frac{{{1^2}}}{2}\log \left| {2 \times 2 + \sqrt {{{(2 \times 2)}^2} + {1^2}} } \right| - \frac{{2 \times 0}}{2}\sqrt {{{(2 \times 0)}^2} + {1^2}} + \frac{{{1^2}}}{2}\log \left| {2 \times 0 + \sqrt {{{(2 \times 0)}^2} + {1^2}} } \right|\]$

$\[\begin{gathered} \int_0^2 {\sqrt {{{(2x)}^2} + {1^2}} } dx = 2\sqrt {17} + \frac{1}{2}\log \left| {4 + \sqrt {17} } \right| \hfill \\ \int_0^2 {\sqrt {{{(2x)}^2} + {1^2}} } dx = 9.156units \hfill \\ \end{gathered} \]$

Therefore, the length of the given curve is 9.156 units.