Posted by **Anonymous** on Friday, December 21, 2012 at 1:16pm.

If 0 <= k <= pi/2 and the area of the region in the first quadrant under the graph of y = 2x-sinx from 0 to k is 0.1, then k =

(a) 0.444

(b) 0.623

(c) 0.883

(d) 1.062

(e) 1.571

- Calculus -
**Steve**, Friday, December 21, 2012 at 2:49pm
a = ∫[0,k] 2x-sinx dx

= x^2+cosx [0,k]

= (k^2+cos(k))-(1)

= cos(k) + k^2-1

so, cos(k) + k^2-1 = 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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