Add maths integration question
The question is "Find f(x) if you are told that f'(x)=10x^4+6x^2+1 and f(1) = 3
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
Then to find C, I tried putting 1 through what I integrated without C to get 4.25. I know that f(1) = 3, so I did 4.25 - 3 = 1.25, so I put that C = 1.25.
Is that right? There aren't any answers on the worksheets I'm doing. Thanks!
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
Then to find C, I tried putting 1 through what I integrated without C to get 4.25. I know that f(1) = 3, so I did 4.25 - 3 = 1.25, so I put that C = 1.25.
Is that right? There aren't any answers on the worksheets I'm doing. Thanks!
Original post by Astravolt
The question is "Find f(x) if you are told that f'(x)=10x^4+6x^2+1 and f(1) = 3
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
Then to find C, I tried putting 1 through what I integrated without C to get 4.25. I know that f(1) = 3, so I did 4.25 - 3 = 1.25, so I put that C = 1.25.
Is that right? There aren't any answers on the worksheets I'm doing. Thanks!
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
Then to find C, I tried putting 1 through what I integrated without C to get 4.25. I know that f(1) = 3, so I did 4.25 - 3 = 1.25, so I put that C = 1.25.
Is that right? There aren't any answers on the worksheets I'm doing. Thanks!
You have the right idea, but you have not integrated the "+ 1" piece correctly. What would you differentiate to end up just with 1?
Think you made a mistake when integrating it, there shouldn't be a 0.5^2 as 1 integrated is just x.
Then just put 1 into f(x) -> 2 + 2 + 1 + C = 3, solve for C.
Then just put 1 into f(x) -> 2 + 2 + 1 + C = 3, solve for C.
Original post by Astravolt
f'(x)=10x^4+6x^2+1
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
I answered it by first integrating to get: 2x^5 + 2x^3 + 0.5^2 + C
? You seem to have integrated a constant (1) as it were the function x. To integrate a constant you just multiply it by x.
To check whether you have integrated correctly you should make sure that you get the original expression when you differentiate your answer.
A constant integrates to just that variable.
The power rule that you're using right now says to add one to exponent and divide by new power.
E.g. integrating 5 dx --> we know 5 = 5x^0 so increasing it becomes 5x/1 = 5x. Apart from that, you're correct.
The power rule that you're using right now says to add one to exponent and divide by new power.
E.g. integrating 5 dx --> we know 5 = 5x^0 so increasing it becomes 5x/1 = 5x. Apart from that, you're correct.
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