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Answer» Applying property `int_(a)^(b)f(a+b-x)dx=int_(a)^(b)f(x)dx` `I_(1)=int_(pi//6)^(pi//3)sec^(2)((pi)/2-theta)f(2sin2((pi)/2-theta))d theta` `=int_(pi//6)^(pi//3)cosec^(3) theta f(2sin 2 theta) d theta=I_(2)` b. `f(x+1)=f(x+3)` or `f(x)=f(x+2)` Thus, `f(x)` is periodic with period 2. Then `int_(a)^(a+b)f(x)dx` is independent of `a` for which `b` is multiple of 2. Thus, `b=2,4,6`........... c. Let `I=int_(1)^(4)(tan^(-1)[x^(2)])/(tan^(-1)[x^(2)]+tan^(-1)[25+x^(2)-10x])`.............1 Appling `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`, we get `I=int_(1)^(4)(tan[(5-x)^(2)])/(tan^(-1)[(5-x)^(2)]+tan^(-1)[x^(2)])dx`.............2 Adding equatiions 1 and 2 we get `2I=int_(1)^(4)dx` or `2I=3` or `I=3//2` d.Let` y=SQRT(x+sqrt(x+sqrt(x+)))..............=sqrt(x+y)` or `y^(2)-y-x=0` or `y=(1+-sqrt(1+4x))/(2.1)=(1+sqrt(1+4x))/2( :' YGT1)` `:.I=int_(0)^(2)(1+sqrt(1+4x))/2 dx` `=[x/2+((1+4x)^(3//2))/(3/2xx2xx4)]_(0)^(2)` `=[(1+27/12)=(0+1/12)]` `=1+26/12=19/6` `:. [I]=3` `[I]=3` |
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