Calculus II
posted by Scott .
Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2

well, the first question is do they cross?
xsinx=x
sinx=1
x=PI/2
so no crossing.
Then area is
int (xxsinx)dx over limits.
x^2/2 sinx+xcosx
PI/2)^2/21 check that. 
I so happens that the two functions intersect at
x = 0 and x = π/2, the domain of our area
area
= [integral] (xsinx  x)dx from 0 to π/2
= sinx  xcosx  (1/2)x^2  from 0 to π/2
= sinπ/2  π/2(cosπ/2)  π^2/8  (sin0  0  0)
= 1  0  π^2/8
= 1  π^2/8
=  .2337
OOPS, just realized that my assumption that the trig curve was above the straight line was false, so we have to reverse the integrand
area = integral (x  xsinx)dx
make the necessary changes, only the signs will be affected, or
we could just take the absolute value of each of my lines.
Respond to this Question
Similar Questions

Calc II
Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi divided by 2 
Calculus 3
Compute the average value of following fuction over the region R? 
math
Find the area of the region between the curves y = sin x and y = x^2  x, 0 ≤ x ≤ 2. 
algebra 1 help please
4) a student score is 83 and 91 on her first two quizzes. write and solve a compound inequality to find possible values for a thord quiz score that would give anverage between 85 and 90. a. 85≤83+91+n/3 ≤90; 81≤n≤96 … 
CALCULUS
Sketch the region enclosed by the given curves. y = tan 3x, y = 2 sin 3x, −π/9 ≤ x ≤ π/9 then then find the area. i can sketch but cant find correct area 
PRE  CALCULUS
Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations. x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π A. x2  y2 = 6; 6 ≤ x ≤ 6 B. x2  y2 = 36; 6 ≤ … 
math
Find the area of the region. (Round your answer to three decimal places.) between y = cos t and y = sin t for −π/2 ≤ t ≤ π/2 
Calculus III
Evaluate ∭W f(x,y,z)dV for the function f and region W specified: f(x,y,z) = 18(x+y) W: y≤ z≤x ; 0≤y≤x ; 0≤x≤1 ∭W (18(x+y))dV = 
Differentials (calc)
Solve the Poisson equation ∇^2u = sin(πx) for 0 ≤ x ≤ 1and 0 ≤ y ≤ 1 with boundary conditions u(x, 0) = x for 0 ≤ x ≤ 1/2, u(x, 0) = 1 − x for 1/2 ≤ x ≤ 1 and 0 everywhere … 
Calculus
Use the limit process to find the area of the region between the graph of the function and the yaxis over the given yinterval. f(y) = 7y, 0 ≤ y ≤ 2