Calculus
posted by Anonymous .
If 0 <= k <= pi/2 and the area of the region in the first quadrant under the graph of y = 2xsinx from 0 to k is 0.1, then k =
(a) 0.444
(b) 0.623
(c) 0.883
(d) 1.062
(e) 1.571

a = ∫[0,k] 2xsinx dx
= x^2+cosx [0,k]
= (k^2+cos(k))(1)
= cos(k) + k^21
so, cos(k) + k^21 = 0.1
k must be small, so cos(k)1 is near 0.
In fact, since cos(k) = 1  k^2/2 + k^4/4!  ...
a =~ k^2/2
so, k =~ √0.2 = 0.44
so, (a)
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