Integration help

[1141]

I've started doing Integration and I need some help with two questions.

" a.) The graph of y = f(x) passes through (4,25) and f'(x) = 6?x. Find its equation.
b.) A curve passes through the point (25,3) and is such that dy/dx = 1/(2?x). Find the equation of the curve."

Can anyone help?

 

[Deleted member 4993]

I've started doing Integration and I need some help with two questions.

" a.) The graph of y = f(x) passes through (4,25) and f'(x) = 6?x. Find its equation.
b.) A curve passes through the point (25,3) and is such that dy/dx = 1/(2?x). Find the equation of the curve."

Can anyone help?

Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you.

To start - find the anti-derivative of f'(x) = 6?x

 

[1141]

To start - find the anti-derivative of f'(x) = 6?x

I think the anti-derivative would be :

6(1 / 3/2 x[sup:b4a0aetm]3/2[/sup:b4a0aetm]) + k
= 6(2/3 x[sup:b4a0aetm]3/2[/sup:b4a0aetm]) + k
= 4x[sup:b4a0aetm]3/2[/sup:b4a0aetm] + k

? y = 4x[sup:b4a0aetm]3/2[/sup:b4a0aetm] + k

Then I would need to solve for k:

y - 4x[sup:b4a0aetm]3/2[/sup:b4a0aetm] = k
25 - 4(4)[sup:b4a0aetm]3/2[/sup:b4a0aetm] = k
? -7 = k

so the equation would be:

y = 4x[sup:b4a0aetm]3/2[/sup:b4a0aetm] - 7

That is the answer I got to. But in my text book they say the answer is y = 4x?x - 7. Is there a difference between my answer and their answer??

 

[Deleted member 4993]

To start - find the anti-derivative of f'(x) = 6?x

I think the anti-derivative would be :

6(1 / 3/2 x[sup:r45dlph8]3/2[/sup:r45dlph8]) + k
= 6(2/3 x[sup:r45dlph8]3/2[/sup:r45dlph8]) + k
= 4x[sup:r45dlph8]3/2[/sup:r45dlph8] + k

? y = 4x[sup:r45dlph8]3/2[/sup:r45dlph8] + k

Then I would need to solve for k:

y - 4x[sup:r45dlph8]3/2[/sup:r45dlph8] = k
25 - 4(4)[sup:r45dlph8]3/2[/sup:r45dlph8] = k
? -7 = k

so the equation would be:

y = 4x[sup:r45dlph8]3/2[/sup:r45dlph8] - 7

That is the answer I got to. But in my text book they say the answer is y = 4x?x - 7. Is there a difference between my answer and their answer?? ... NO

4x?x = 4 * x * ?x = 4 * x[sup:r45dlph8]1[/sup:r45dlph8] * x[sup:r45dlph8]1/2[/sup:r45dlph8] = 4 * x[sup:r45dlph8](1+ 1/2)[/sup:r45dlph8] = 4 * x[sup:r45dlph8]3/2[/sup:r45dlph8]

 

[1141]

Alright. Thank you.

And I tried question b.) but I get the wrong answer. Can you tell me what I'm doing wrong?

f'(x) = 1 / (2?x)
f(x) = 2x[sup:ecjknq5m]1/2[/sup:ecjknq5m]
= 2(1 / -3/2 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m]) + k
= 2(-2/3 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m]) + k
= -4/3 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m] + k

? y = -4/3 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m] + k

Then I solve for k,

y + 4/3 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m] = k
5 + 4/3 (1/2)[sup:ecjknq5m]-3/2[/sup:ecjknq5m] = k
8.771236... = k

? y = -4/3 x[sup:ecjknq5m]-3/2[/sup:ecjknq5m] + 8.77

 

[wjm11]

b.) A curve passes through the point (25,3) and is such that dy/dx = 1/(2?x). Find the equation of the curve."

f'(x) = 1 / (2?x)
f(x) = 2x1/2

f'(x) = 1 / (2?x) = (1/2)(x^-.5)
f(x) = x^.5 + k
3 = 25^.5 + k
k = 3 – 5 = -2