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AAKASH SERIES-DEFINITE INTEGRALS-EXERCISE - I
- {:(" "Lt),(n rarr oo):}1/n sum(r=1)^(2n)(r)/(sqrt(n^(2)+r^(2)))=
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- {:(" "Lt),(n rarr oo):}1/n { f(1/n)+f(2/n)+...+f(2)}=
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- {:(" "Lt),(n rarr oo):}(1)/(n^(6)){(n+1)^(5)+(n+2)^(5)+...+(2n)^(5)}=
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- {:(" "Lt),(n rarr oo):} 1/n [Sin^(2). (pi)/(2n)+Sin^(2). (2pi)/(2n)+.....
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- {:(" "Lt),(n rarr oo):} [(1)/(1-n^(2))+(2)/(1-n^(2))+...+(n)/(1-n^(2))...
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- Lt(ntooo)[(1)/(sqrt(n^(2)-1^(2)))+(1)/(sqrt(n^(2)-2^(2)))+(1)/(sqrt((2...
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- If int(0)^(k)(1)/(2+8x^(2))dx=(pi)/(16) then k=
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- The value of I=int(0)^(pi//2)((sinx+cosx)^(2))/(sqrt(1+sin2x))dx is
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- int(0)^(pi)(1)/(1+sinx)dx=
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- int(0)^(pi//4)(Sin^(9)x)/(Cos^(11)x)dx=
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- int(0)^(1)sqrt((1-x)/(1+x))dx=
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- int(2)^(5) sqrt((x-2)/(5-x))dx=
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- int(2)^(3) (dx)/(sqrt((x-2)(3-x)))=
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- int(0)^(1)sqrt(x(1-x))dx=
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- int(0)^(pi//2)(1)/(4 cos^(2)x+9sin^(2)x)dx=
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- int(0)^(pi//4) (dx)/(2+sin^(2)x)=
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- int(0)^(pi//2) x sin x dx =
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- int(0)^(pi//2) e^(x) (cos x - sin x ) dx =
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- int(0)^(1)(xe^(x))/((x+1)^(2))dx=
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- int(0)^(1) x sin^(-1) x dx =
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