Math2400
  • Math2400
can someone explain this? I don't understand my instructors steps:
Mathematics
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SOLVED
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katieb
  • katieb
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Math2400
  • Math2400
suppose that f(1) =2. f(4)=5, f'(4)=3, and f'' is continuous. Find the value of \[\int\limits_{1}^{4} xf''(x)dx\]
Math2400
  • Math2400
the answer is 9 but im not sure how my prof. did it. i think he tried taking the anti-deriv. first. or used some kind of property. cuz he got something like: |dw:1433615204902:dw|
Math2400
  • Math2400
but idk how he got that>< can anyone explain?

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Math2400
  • Math2400
@Luigi0210 would u be able help?
Math2400
  • Math2400
@ganeshie8 another one if you're free :) i'm studying for my final><
amistre64
  • amistre64
looks like a by parts application
Luigi0210
  • Luigi0210
^ -.-
amistre64
  • amistre64
what is the derivative of a product?
amistre64
  • amistre64
lets take the derivative of the product of 2 function, f and g [fg]' = ??
amistre64
  • amistre64
or, to better line up with your specifics .. [gf]' = ??
Math2400
  • Math2400
g'f+gf'
amistre64
  • amistre64
good [gf]' = g'f + gf' now integrate both sides
amistre64
  • amistre64
[gf]' integrates back in gf and we have the sum of integration on the right \[gf=\int g'f~+~\int gf'\] now, we want to know how to take the integration of gf', so lets subtract to get it all by itself ... and we end up with the by parts formulation \[\int gf'=gf-\int g'f\]
amistre64
  • amistre64
let g=x, and f' = f''
amistre64
  • amistre64
the letters we named our function are immaterial, one usual notation is to use u and v \[\int uv'=uv-\int vu'\]
ybarrap
  • ybarrap
I think that you will also need \(f'(1)=0\) as a \(\bf{given}\) later in this process.
phi
  • phi
Hopefully you have leaned integration by parts (which is what Amistre was deriving) https://en.wikipedia.org/wiki/Integration_by_parts You use that "rule" to get your result
phi
  • phi
you also need to now that \[ \int \frac{d^2 f(x)}{dx^2} \ dx = \frac{d f(x)}{dx} = f'(x)\]
phi
  • phi
*know
ybarrap
  • ybarrap
Applying the rules of @amistre64 we have $$ [xf']'=x'f'+xf''=f'+xf'' $$ Then $$ \int [xf']'dx=xf'\\ =\int \left (x'f'+xf''\right ) dx\\ =\int \left (f'+xf''\right ) dx\\ $$ From which you get what you have written above. Now evaluate. You are missing an initial condition: \(f'(1)=0\). Using this, you'll get 9 after evaluation.

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