The required value of [tex]\int\limits^4_2 {f(x)g"(x)} \, dx (1)[/tex] is 63.
Given that,
The table above gives selected values of twice-differentiable functions f and g,
The first two derivatives of g. If f′ (x) = 3 for all values of x.
We have to determine,
What is the value of [tex]\int\limits^4_2 {f(x)g"(x)} \, dx (1)[/tex].
According to the question,
To obtain the value of given function, integrating the function,
[tex]= \int\limits^4_2 {f(x)g"(x)} \, dx (1)[/tex]
Integrating by using by-parts.
[tex]= \int\limits^4_2 {f(x)g"(x)} \, dx (1)\\\\if\ u = f(x), \ then \ du = f'(x)dx\\\\if \ dv = g"(x)dx, \ then \ v = g'(x)\\\\[/tex]
Then,
[tex]\int u\ dv = uv\ - \ \int v \ du\\\\= f(x) \ g'(x) - \int g'(x) f'(x) \ dx\\[/tex]
if f'(x) is constant at 3,
Then,
[tex]= f(x) g'(x) - \int g'(x)\dx\\\\= f(x) g'(x) -3g(x)[/tex]
Substitute from x = 2 to x = 4 then evaluate,
[tex]= [f(4)\ g'(4) - 3 \ g (4)]- [f(2)\ g'(2) - 3 \ g (2)]\\\\= [(13).(7) - 3.(9)] - [7.(1)- 3.(2)]\\\\= [91-27} - {7-6}\\\\= 64-1\\\\= 63[/tex]
Hence, The required value of [tex]\int\limits^4_2 {f(x)g"(x)} \, dx (1)[/tex] is 63.
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