The value of ∫xf'(x)dx with upper and lower limits 0 to 4 is -20, option (a) is correct.
What is integration?
It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.
We have a function f which has a continuous derivative and the value of
[tex]\rm \int\limits^4_0 {f(x)} \, dx =8[/tex]
To find the value of
[tex]\rm \int\limits^4_0 {x \ f'(x)} \, dx[/tex]
We will use integration by parts.
Let's suppose u = x and v = f'(x)
We know:
[tex]\rm \int u \ v \ dx =u\int v dx -\int u'(\int vdx)dx[/tex]
[tex]\rm \int x \ f'(x) \ dx =x\int f'(x) dx -\int 1(\int f'(x)dx)dx[/tex] (u' = 1)
[tex]\rm \int x \ f'(x) \ dx =xf(x) -\int f(x)dx[/tex] [tex](\rm \int f'(x) dx = f(x))[/tex]
After applying limits, we get:
[tex]\rm \int\limits^4_0 {xf'(x)}dx \, =|xf(x)|^4_0 - \int\limits^4_0 {f(x)} \, dx[/tex]
[tex]\rm \int\limits^4_0 {xf'(x)}dx \, = [ 4f(4)-0f(0)] - \int\limits^4_0 {f(x)} \, dx[/tex]
From the table f(4) = -3 and f(0) 2 put these values in the above expression, we get
[tex]\rm \int\limits^4_0 {xf'(x)}dx \, = [ 4(-3)] - 8 \Rightarrow -20[/tex] [tex]( \rm \int\limits^4_0 {f(x)} \, dx =8)[/tex]
Thus, the value of ∫xf'(x)dx with upper and lower limits 0 to 4 is -20, option (a) is correct.
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