PYQSidePowered by
ExamsMy practice
x 2 ( x3( /2)3 ( (2t1/3 ) + (t1/3 ) ) dt (x - 2 )2 ) is equal to
(2) 9 28
( 00 ) form; by L'Hospital's rule and Leibnitz rule, ddx x3( /2)3 f(t) dt = -3x2 f(x3):
L = x 2 -3x2( 2x + x)2 (x - 2 )
At x = 2, 2x + x = 0, so it is again 00; apply the rule once more (the term with ( 2x + x) vanishes):
L = -3x2(2 2x - x)2 at x = 2
= -3 24(-2 - 1)2 = 9 28
Answer: (2)
The value of the integral -12 e (x + x2 + 1 ) dx is
The value of k N for which the integral In = 01 (1 - xk )n dx, n N, satisfies 147 I20 = 148 I21 is
Let I(x) = 6 2 x(1 - x)2 dx. If I(0) = 3, then I ( 12 ) is equal to
Let e 4 dx ex - 1 = 6. Then e and e- are the roots of the equation :
If 1 [5](x - 1)4(x + 3)6 dx = A ( x - 1 x + 3 )B + C, where C is the constant of integration, then the value of + + 20AB is .
0 /4 2 x 2 x ( 3 x + 3 x )2dx is equal to
Let rk = 01 (1 - x7 )k dx 01 (1 - x7 )k+1 dx, k N. Then the value of k=110 17 (rk - 1 ) is equal to .
If 1a2 2 x + b2 2 x dx = 112 -1(3 x) + constant, then the maximum value of a x + b x, is :
Let [t] denote the largest integer less than or equal to t. If 03 ( [x2 ] + [ x22 ] )dx = a + b 2 - 3 - 5 + c 6 - 7, where a, b, c Z, then a + b + c is equal to .
The integral 0 /4 136 x3 x + 5 x dx is equal to :