The area of the region bounded by the lines `x=0, x=pi/2` and `f(x)=sinx, g(x)=cosx` is (A) `2(sqrt(2)+1)` (B) `sqrt(3)-1` (C) `2(sqrt(3)-1)` (D) `2(sqrt(2)-1)`

The area formed by triangular shaped region bounded by the curves y=sinx, y=cosx and x=0 is (A) sqrt(2)-1 (B) 1 (C) sqrt(2) (D) 1+sqrt(2)

The area of the figure bounded by the curves y=cosx and y=sinx and the ordinates x=0 and x=pi/4 is (A) sqrt(2)-1 (B) sqrt(2)+1 (C) 1/sqrt(2)(sqrt(2)-1) (D) 1/sqrt(2)

The area bounded by the y-axis, y" "=" "cos" "x and y" "=" "s in" "x when 0lt=xlt=pi/2 is (A) 2(sqrt(2-1)) (B) sqrt(2)-1 (C) sqrt(2)+1 (D) sqrt(2)

The area bounded by the curve y=secx , the x-axis and the lines x=0 and x=pi/4 is (A) log(sqrt(2)+1) (B) log(sqrt(2)-1) (C) 1/2log2 (D) sqrt(2)

int_0^(4/pi) (3x^2sin(1/x)-xcos(1/x))dx= (A) (8sqrt(2))/pi^3 (B) (32sqrt(2))/pi^3 (C) (24sqrt(2))/pi^3 (D) sqrt(2048)/pi^3

The area enclosed by the curve y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)] is 4(sqrt(2)-2) (b) 2sqrt(2)(sqrt(2)-1)2(sqrt(2)+1)(d)2sqrt(2)(sqrt(2)+1)

Find the area of the region bounded by the curves y=sqrt(x+2) and y=(1)/(x+1) between the lines x=0 and x=2.

If f(x)=sin^2x and the composite function g(f(x))=|sinx| , then g(x) is equal to sqrt(x-1) (b) sqrt(x) (c) sqrt(x+1) (d) -sqrt(x)

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

If the area of the triangle formed by the lines x=0,y=0,3x+4y-a(a>0) is 1, then a=(A)sqrt(6)(B)2sqrt(6)(C)4sqrt(6)(D)6sqrt(2)