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HIMALAYA PUBLICATION-Definite Integrals and its Application-QUESTION BANK
- int0^(4/pi) (3x^(2) sin (1/x) - x cos (1/x)dx =
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- int0^(sqrtlog (pi/2)) cos(e^(x^(2))).2xe^(x^(2)) dx =
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- int0^(pi/2) (cosx-sinx)/(1+sinx cosx) dx =
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- int(-1/2)^(1/2) [sin^(-1)(3x-4x^(3)) - cos^(-1) (4x^(3)-3x)] dx =
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- int(-2)^(1) cot^(-1) (1/x) dx =
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- I = int(-2)^(1) (tan^(-1)x+ cot^(-1)(1/x) dx
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- int 1^(oo) (e^(x+1)+e^(3-x))^(-1) dx =
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- int1^(2) (dx)/(x(1+x^(4))) =
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- I(n) = int0^(pi/4) tan^(n) x dx, n in N, then I(10)+I(8) =
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- int0^(pi) sin^(50)x.cos^(49)xdx=
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- int sqrtx. e^(sqrtx) dx =
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- int0^(10) x^(10)/((10-x)^((10))+x^(10)) dx =
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- int0^(2) [2x]dx =, where [.] denotes the greatest function.
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- The integral int(-1//2)^(1//2) ([x] + ln ((1 + x)/(1 - x)))dx equals :
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- If [.] stands for the greatest integr function, int1^(2)[3x]dx =
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- int0^(6) |x-3|dx=
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- If f(x) be a continous function for all real values of x and satisfies...
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- If f(x) = int(2x)^(sinx) cos (t^(3)) dt then f '(x)
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- Iff(x) = int1^(x^(2)) tan^(-1) sqrtt dt, t gt 0, then f^(')(1)=
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- If int0^x f(t) dt = x + intx^l tf (t) dt, then the value of f(1) is :
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