Calculus
posted by Michelle .
Find the area of the region enclosed between y = 2sin (x) and y = 2cos (x) from x = 0 to x = pi/4.
Thanks for your help :)

Since cos x > sin x on the interval,
Integrate 2cos x  2sin x on [0,pi/4]
2sin x + 2cos x [0,pi/4]
at x=pi/4, 2(1/√2 + 1/√2) = 2*2/√2 = 2√2
at x=0, 2*0 + 2*1 = 2
area = 2√2  2
Respond to this Question
Similar Questions

calculus
find the value of m so that the line y = mx divides the region enclosed by y = 2xx^2 and the xaxis into two regions with equal area i know that the area enclosed by y=2xx^2 is 4/3 so half of that would be 2/3 however, i have no … 
Calculus
Determine the area of the region enclosed by the limaçon r=5+2sin(theta) 
Calc 2: Area under the curve
Find the area of the region enclosed between y=2sin(x and y=3cos(x) from x=0 to x=0.4pi Hint: Notice that this region consists of two parts. Notice: I'm getting 1.73762 but apparently that is wrong. 
Calculus
Find the area of the region enclosed by the given curves: y=e^6x, y=2sin(x), x=0, x=pi/2 
calculus
the region enclosed by 2y=5xy=5 and 2y+3x=8. Decide whether to integrate with respect to x or y, and then find the area of the region. The area is 
CalculusArea between curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even … 
Calculus Area between curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2x=1 
calculus
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=3+√X, y=3+1/5x What is the area? 
Calculus
Any help will be appreciated! Find the area of the region between x=3pi/4 and 5pi/4 in the graph enclosed by f(x)=sinx and g(x)=cosx Thank you! 
Calculus
Find the area of the region enclosed between y=4sin(x) and y=2cos(x) from x=0 to x=0.7pi. Hint: Notice that this region consists of two parts.