find the area between y=sinx and y=3cosx over the interval (0,pi) Solution PLEAS

Question

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU It helps to graph the two curves. 2sin(x) starts at (0,0) and rises to (pi/2, 2). 3cos(x) starts at (0, 3) and decreases to (pi/2,0) Therefore, these curves intersect inside the interval of interest. It occurs when 2sin x = 3cos x which means x = arctan(3/2) = approx .98. To the left of the intersection, 3cos x is above 2sin x; To the right of the intersection, 2sin x is above 3cos x. The area between 2 curves that are given by f and g WHERE f is greater than or equal to g is given by the integral of f - g. In this problem, on the interval x=0 to arctan(3/2) , the area bounded by the 2 curves is therefore given by the integral of (3cos(x) - 2sin(x) ) {since 3cos is > 2sin on this interval} On the interval x=arctan(3/2) to x = pi/2, the area bounded by these two curves is given by the integral of ( 2sin(x) - 3cos(x) ) {since 2sin is > 3cos on this interval} So the total area bounded by the 2 curves on the interval [0, pi/2] is given by the sum of the two integrals. I get a result of (after combining and simplifying, etc): 6sin[ arctan(3/2) ] + 4cos[ arctan(3/2) ] - 5 which is approx. 2.211 I checked this on a graphing calculator and got the same numerical result.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.