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dusya [7]
3 years ago
13

What is the integral of a square root? Please explain.

Mathematics
2 answers:
dolphi86 [110]3 years ago
6 0
Its not guaranteed to be same form.it could be any polynomial that can not be easily formed into squares .try one out!
IrinaK [193]3 years ago
5 0
If y = xⁿ

∫y dx =  xⁿ⁺¹ / (n + 1)  +  C    Provided n ≠ -1.

y = √x

y = x^(0.5)

∫y dx  =  x^(0.5+1) / (0.5 + 1) =   x^(1.5)  / 1.5 = x^(1.5)  / (3/2)

∫y dx = (2/3) x^(1.5)  + C.

∫y dx = (2/3) x^(3/2)  + C.

∫y dx = (2/3)√x³  + C
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The value of
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\large\underline{\sf{Solution-}}

We have to find out the value of the fraction.

<u>Let us assume that:</u>

\sf \longmapsto x =2 +   \dfrac{1}{2 +  \dfrac{1}{2 +  \dfrac{1}{2 + ... \infty} } }

<u>We can also write it as:</u>

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<u>Comparing </u>the given <u>equation</u> with <u>ax² + bx + c = 0,</u> we get:

\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b =  - 2 \\ \sf c =  - 1 \end{cases}

<u>By quadratic formula:</u>

\sf \longmapsto x =  \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{8} }{2}

\sf \longmapsto x =  \dfrac{2 \pm2 \sqrt{2} }{2}

\sf \longmapsto x = 1 \pm\sqrt{2}

\sf \longmapsto x = \begin{cases} \sf 1  + \sqrt{2} \\ \sf 1 -  \sqrt{2}  \end{cases}

<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>

\sf :\implies x = 1 + \sqrt{2}

So, the value of the fraction is 1 + √2.

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