Let I(x)= dx(x-11) 1113(x+15) 1513. If I(37)-I(24)= 14 ( 1b 113- 1c 113 ), b,c N, then 3(b+c) is equal to
(1) 39
Powers add up to 1113+ 1513=2 put t= x-11x+15, dt= 26 dx(x+15)2
I= 126 t-11/13dt= 14t2/13
t(37)= 12, t(24)= 13
I(37)-I(24)= 14 [ 141/13- 191/13 ] =4, c=9
3(b+c)=39
Answer: (1)
If I= 0 2 32x 32x+ 32x dx, then 02I x x x 4x+ 4x dx equals :
Let x3 x dx=g(x)+C, where C is the constant of integration. If 8 (g ( 2 )+g' ( 2 ) )= 3+ 2+ , , , Z, then + - equals :
Let for f(x)=7 8x+7 6x-3 4x-3 2x, I1= 0 /4f(x) dx and I2= 0 /4x f(x) dx. Then 7I1+12I2 is equal to :
If x ( x -1x 1-x2+ -1x(1-x2)3/2+ x1-x2 )dx=g(x)+C, where C is the constant of integration, then g ( 12 ) equals :
Let 2 - x3 + x dx = 12 ( x + e| x + x| ) + C, where C is the constant of integration. Then + is equal to :
Let n ( n n4 + 1 - 2n (n2 + 1 ) n4 + 1 . . + n n4 + 16 - 8n (n2 + 4 ) n4 + 16 + . . + n n4 + n4 - 2n n2 (n2 + n2 ) n4 + n4 ) be k, using only the principal values of the inverse trigonometric functions. Then k2 is equal to .
The integral 1/43/4 (2 -1 1 - x1 + x ) dx is equal to
x 2 ( x3( /2)3 ( (2t1/3 ) + (t1/3 ) ) dt (x - 2 )2 ) is equal to
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