Homework Help: Calculus

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)

Answer this Question

Related Questions

Calculus - Let R be the region in the first quadrant under the graph of y=1/sqrt...
calculus - given the graph of f(x) = x sinx, 0<=x<=2pi assuming ...
calculus - Let A be the area if the region in the first quadrant under the graph...
Calculus - I am doing the AP calculus review, these are the questions I have no ...
calculus - find the area of the region under y=6 ln(3 x) and above y=8 for 4&...
C++ Programming - Can any1 please fix this c++ source problem /*Circle Are ...
Calculus Help!! - Region R is bounded by the functions f(x) = 2(x-4) + pi, g(x...
calculus - The region R is defined by 1(</=)x(</=)2 and 0(&lt...
Calculus - Find the area under the curve y=5(cosx) and above y=5(sinx) for 0&...
Calculus - Find the area of the region which is bounded by the polar curves ...

For Further Reading

First Name:
School Subject:
Answer: