[SOLVED] Help needed with simple integration!

Helonicus · January 5th 2008, 05:40 AM

Hi people,

Can anyone help me at all?
I have a couple of questions regarding intergration and differentiation that i am stuck on.

1) Differentiation question:-

Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
is equal to a) -1
b) 0
c) 2
I started this and got to dy/dx = 4-2x then stuck help!

2) Integration question:-

Integrate with respect to 'x'

a) 1 divided by the square root of 'x' (1/sqrt 'x')

Please feel free to explain your answers as i would like to learn/understand the method of solving involved.

Many thanks

**earboth** · January 5th 2008, 06:27 AM

Originally Posted by Helonicus

1) Differentiation question:-

Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
is equal to a) -1
b) 0
c) 2
I started this and got to dy/dx = 4-2x correct!

2) Integration question:-

Integrate with respect to 'x'

a) 1 divided by the square root of 'x' (1/sqrt 'x')

Hello,

to 1).

You got the derivative correctly. You are asked to calculate the value of x if

4-2x = -1..................(5/2)
4-2x = 0 .................. (2)
4-2x = 2 .................. (1)

to 2):

$\frac1{\sqrt{x}} = x^{-\frac12}$

So you have to calculate:

$\int\left(x^{-\frac12} \right) dx = \frac1{\frac12} \cdot x^{-\frac12 + 1}+C = 2 \cdot \sqrt{x} + C$

**polymerase** · January 5th 2008, 06:38 AM

Originally Posted by Helonicus

Hi people,

Can anyone help me at all?
I have a couple of questions regarding intergration and differentiation that i am stuck on.

1) Differentiation question:-

Find the values of 'x' for which the gradient of the curve y=3+4x-x^2
is equal to a) -1
b) 0
c) 2
I started this and got to dy/dx = 4-2x then stuck help!

2) Integration question:-

Integrate with respect to 'x'

a) 1 divided by the square root of 'x' (1/sqrt 'x')

Please feel free to explain your answers as i would like to learn/understand the method of solving involved.

Many thanks

Hey,

ok so the gradient of the curve at the points you mention just mean to find the tangent line at those line. To find tangent line, you use its derivative.
So, $f(x)=3+4x-x^2$ , $f'(x)=4-2x$ .
Then you just set the derivative equal to each of the three points and so for x. You should get $x=\frac{5}{2},2,1$ .

$\int \frac{1}{\sqrt{x}}$ . Let $x=sin^2u$ , then $du=2sin\;u\;cos\;u$ .

$2\int \frac{sin\;u\;cos\;u}{sin\;u}du=2\int cos\;u=2sin\;u$ . Back sub and your done.

Hence, $\int \frac{1}{\sqrt{x}}=2\sqrt{x}$

**Krizalid** · January 5th 2008, 06:52 AM

Originally Posted by polymerase

$\int \frac{1}{\sqrt{x}}$ . Let $x=sin^2u$ , then $du=2sin\;u\;cos\;u$ .

$2\int \frac{sin\;u\;cos\;u}{sin\;u}du=2\int cos\;u=2sin\;u$ . Back sub and your done.

Hence, $\int \frac{1}{\sqrt{x}}=2\sqrt{x}$

Ohh c'mon!!

This is rekillin' a fly with a Huge Grand Cannon!

Even it's a valid solution, power rule it's faster and easy to handle.

--

Remember the title "simple integration", we don't actually know if Helonicus handles trigonometric substitution.

**polymerase** · January 5th 2008, 09:22 AM

Originally Posted by Krizalid

Ohh c'mon!!

I know its the long way of do it....but earboth already did it the power way...besides its only long because i wrote out the entire thing for him/her to see how do it....i would've done it in my head.

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