Question

Find f. Antiderivatives.

Answer 1

#11

Answer 2

Gonna need some paper for this one! Bear with me

Answer 3

\[f ' (x) =4x+sinx+C\]
\[f(x)=2x^2-cosx+Cx+D\]
\[f(0) =2(o)^2-\cos(0)+C(o)+D=-1\]
\[0-1+O+D=-1\]
\[D=0\]
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\[f(7\pi/2)=2(7\pi/2)^2-\cos(7\pi/2)+(7\pi/2)C+D=0\]
\[2\frac{ 14\pi^2 }{ 4 }-0+(7\pi/2)C=0\]
\[\frac{ 28\pi^2 }{ 4 }-0+(7\pi/2)C=0\]
\[7\pi^2+(7\pi/2)C=0\]
\[(7\pi/2)C=-7\pi^2\]
\[C=-\frac{ 7\pi^2 }{ 2 }\times \frac{ 2 }{ 7\pi }\]
\[C=-7\pi\]
\[f(x)=2x^2-cosx-7\pi\]

Answer 4

Here is my attempt, but the answer is wrong.

Answer 5

ops, 49pi/4*

Answer 6

Aww so close man, you're going to kick yourself for this one. Your mistake is on the last line, it should be:\[f(x)=2x^2-cos(x)-7\pi x\]

Answer 7

I'm going to redo the problem, 2(7pi)/2)^2= is 98pi/4

Answer 8

What you did was right, you just missed the x on the end. f(x)=2x^2−cosx+Cx+D.

Answer 9

isn't \[2\times(\frac{ 7\pi }{ 2 })^2=\frac{ 98\pi^2 }{ 4 }\] and not \[\frac{ 28\pi^2 }{ 4 }\]?

Answer 10

@tom982

Answer 11

okay, even with (98pi^2/4), it is still -7pi

Answer 12

I see the answer now, thanks! :0

Answer 13

I was redoing the problem by using 98pi^2/4 instead of 28pi^2/4. So, it took me a while to respond back. Hehe

Answer 14

Maybe my working will help:
\[f''(x)=4+\cos(x)\]\[f'(x)=4x+\sin(x)+c\]\[f(x)=2x^2-\cos(x)+cx+d\]
\[f(0)=-1=0-1+0+d \]\[d=0\]
\[f(\dfrac{7\pi}{2})=0=2(\dfrac{7\pi}{2})^2-cos(\dfrac{7\pi}{2})+\dfrac{7\pi}{2}c+d\]\[0=2(\dfrac{7\pi}{2})^2+\dfrac{7\pi}{2}c\]\[2(\dfrac{7\pi}{2})^2=-\dfrac{7\pi}{2}c\]\[2\dfrac{7\pi}{2}=-c\]\[c=2\dfrac{7\pi}{2}=7\pi \]

Looking back at your work, you made a mistake and accidentally found the right answer:
\[(7\pi/2)C=-7\pi^2\]\[C=-\frac{ 7\pi^2 }{ 2 }\times \frac{ 2 }{ 7\pi } \]
You've divided by 2 out of nowhere.