InterviewSolution
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InterviewSolution
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Evaluate : `(i) int_(0)^(2)e^(x//2)dx` `(ii) int_(2)^(4)(x)/((x^(2)+1))dx` `(iii) int_(0)^(1)cos^(-1)xdx` |
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Answer» `(i)` Put `(x)/(2)=t` so that `dx=2dt`. Also, `(x=0impliest=0)` and `(x=2=t=1)`. `:.int_(0)^(2)e^(x//2)dx=2int_(0)^(1)e^(t)dt=2[e^(t)]_(0)^(1)=2(e-1)`. `(ii)` Put `(x^(2)+1)=t` so that `xdx=(1)/(2)dt`. Also, `(x=2impliest=5)` and `(x=4impliest=17)`. `:.int_(2)^(4)(x)/((x^(2)+1))dx=(1)/(2)int_(5)^(17)(dt)/(t)=(1)/(2)[log|t|]_(5)^(17)=(1)/(2)(log17-log5)`. `(iii)` Put `x=cost` so that `dx=-sint dt`. Also, `(x=0impliest=(pi)/(2))` and `(x=1impliest=0)`. `:.int_(0)^(1)cos^(-1)xdx=-int_(pi//2)^(0)cos^(-1)(cost)sintdt=int_(0)^(pi//2)tsintdt` `=[t(-cost)]_(0)^(pi//2)-int_(0)^(pi//2)1*(-cost)dt` [integrating by parts] `=[sint]_(0)^(pi//2)=1`. `(iv)` Let `(2x+3)-=A*(d)/(dx)(5x^(2)+1)+B`. Then, `(2x+3)-=(10x)A+B`. Comparing the coefficients of like powers of `x`, we get `10A=2` or `A=(1)/(5)` and `B=3`. `:.(2x+3)=(1)/(5)(10x)+3`. So, `int_(0)^(1)((2x+3))/((5x^(2)+1))dx=int_(0)^(1)((1)/(5)(10x)+3)/((5x^(2)+1))dx` `=(1)/(5)(10x)/((5x^(2)+1))dx+3int_(0)^(1)(dx)/((5x^(2)+1))` `=(1)/(5)[log|5x^(2)+1|]_(0)^(1)+(3)/(5)int_(0)^(1)(dx)/(x^(2)+((1)/(sqrt(5)))^(2))` `=(1)/(5)log6+(3)/(5)*sqrt(5)[tan^(-1)(x)/((1//sqrt(5)))]_(0)^(1)` `=(1)/(5)log6+(3)/(sqrt(5))(tan^(-1)sqrt(5))`. |
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