How do u integrate this? (1 Viewer)

Kingportable · Dec 4, 2011

How do you integrate this? I'm confused about how to take out the "d".

$\int_{1}^{9}\frac{dx}{\sqrt{x}}$

SpiralFlex · Dec 4, 2011

$=\int_{0}^{9} x^{-\frac{1}{2}} \:\mathrm{d} x$

Add one power, your new power is $\frac{1}{2}$ . Divide by that new power.

$=\frac{x^\frac{1}{2}}{\frac{1}{2}}$

$=2\sqrt{x}$ (With the limits 0 to 9.)

$=2[3-1]$

$=4$

nightweaver066 · Dec 4, 2011

Kingportable said:
How do you integrate this? I'm confused about how to take out the "d".

$\int_{1}^{9}\frac{dx}{\sqrt{x}}$

$\int^9_1 \frac{1}{\sqrt{x}}dx$

$=\int^9_1 x^{-\frac{1}{2}}dx$

$=[2x^{\frac{1}{2}}]^9_1$

$=[2 \times \sqrt{9} - 2 \times \sqrt{1}]$

$=6 - 2$

$=4$

stampede · Dec 4, 2011

same thing as the integral of x^-1/2

just move the dx a little to the right, its just asking for the integral of x^-1/2 between x=1 and x=9

you feel me?

Kingportable · Dec 4, 2011

Yes! Thanks i get it now.

zeebobDD · Dec 4, 2011

taking out the d?

when its dx by itsself it means 1

Shadowdude · Dec 4, 2011

zeebobDD said:
taking out the d? when its dx by itsself it means 1

not sure if srs

How do u integrate this? (1 Viewer)

Kingportable

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SpiralFlex

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nightweaver066

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stampede

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Kingportable

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zeebobDD

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Shadowdude

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