It says "find the area of the region bounded by the graph of f(x)=7x sin(x)
and the x-axis on teh interval [0,pi]."
would i do this...?
integral as 0 goes to pi of 7xsin(x) dx ? or would i have to square it?
Yes. And do not square it.
Area = w*L
dA = y*dx = f(x) *dx = 7x*sin(x) *dx
dx goes from x=0 to x=pi
So,
A = (7)INT.(0 --> pi)[x*sin(x)]dx --------------(i)
(i) cannot be integrated yet because x*dx is not the derivative of sin(x).
We use integration by parts.
INT.[u*dv] = u*v -INT.[v*du] --------------***
Let u = x
So, du =dx
And dv = sin(x) *dx
So, v = -cos(x)
Then the integrand of (i) becomes
7(-x*cos(x) -INT[-cos(x)*dx])
7(-x*cos(x) +INT[cos(x)]dx)
The boundaries of dx remain the same, from x=0 to x=pi
So,
A = 7[-x*cos(x) +sinx]|(0 --> pi)
A = 7[{-pi*cos(pi) +sin(pi)} -{-0 +0}]
A = 7[-pi*(-1) +0]
A = 7pi sq.units ------------------answer.