The value of the integral
∫ 1 { x2 + x2013/(x2 + 2IxI +1)}dx
-1
(1) positive
(2) negative
(3) zero
(4) infinite
dx is
kvnmurty:
the question is not clear... x2013 ?? what does it mean ? what is the expression in numerator... x square + x2013 or only x2013 ? use parentheses if required to clearly write the expression
it is x to the power 2013 and x square .. & limit of the integration is from 1 to -1 .. please help me
Answers
Answered by
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I suppose the given expression inside the integral is as below.
![I= \int\limits^{1}_{x=-1} [{x^2+\frac{x^{2013}}{x^2+2|x| +1}}] \, dx \\\\I= \int\limits^{1}_{-1} {x^2} \, dx + \int\limits^{1}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\=\frac{1}{3}[x^3]_{-1}^{1}+\int\limits^{0}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=\frac{2}{3}-\int\limits^{-1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=2/3](https://tex.z-dn.net/?f=I%3D+%5Cint%5Climits%5E%7B1%7D_%7Bx%3D-1%7D+%5B%7Bx%5E2%2B%5Cfrac%7Bx%5E%7B2013%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%5D+%5C%2C+dx+%5C%5C%5C%5CI%3D+%5Cint%5Climits%5E%7B1%7D_%7B-1%7D+%7Bx%5E2%7D+%5C%2C+dx+%2B+%5Cint%5Climits%5E%7B1%7D_%7B-1%7D+%7Bx%2A%5Cfrac%7B%28x%5E2%29%5E%7B1006%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%7D+%5C%2C+dx+%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B3%7D%5Bx%5E3%5D_%7B-1%7D%5E%7B1%7D%2B%5Cint%5Climits%5E%7B0%7D_%7B-1%7D+%7Bx%2A%5Cfrac%7B%28x%5E2%29%5E%7B1006%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%7D+%5C%2C+dx+%2B%5Cint%5Climits%5E%7B1%7D_%7B0%7D+%7Bx%2A%5Cfrac%7B%28x%5E2%29%5E%7B1006%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%7D+%5C%2C+dx+%5C%5C%5C%5CI%3D%5Cfrac%7B2%7D%7B3%7D-%5Cint%5Climits%5E%7B-1%7D_%7B0%7D+%7Bx%2A%5Cfrac%7B%28x%5E2%29%5E%7B1006%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%7D+%5C%2C+dx+%2B%5Cint%5Climits%5E%7B1%7D_%7B0%7D+%7Bx%2A%5Cfrac%7B%28x%5E2%29%5E%7B1006%7D%7D%7Bx%5E2%2B2%7Cx%7C+%2B1%7D%7D%7D+%5C%2C+dx+%5C%5C%5C%5CI%3D2%2F3 "I= \int\limits^{1}_{x=-1} [{x^2+\frac{x^{2013}}{x^2+2|x| +1}}] \, dx \\\\I= \int\limits^{1}_{-1} {x^2} \, dx + \int\limits^{1}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\=\frac{1}{3}[x^3]_{-1}^{1}+\int\limits^{0}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=\frac{2}{3}-\int\limits^{-1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=2/3")
Let f(x) be function inside ONE Integral on the RHS. Then let X= -x.
In the given integral
is anti symmetric function. so its image wrt y axis. So G(x) = - G(-x). Hence, the area under the curve G(x) for x > 0 is equal and opposite to the area under the curve for x < 0.
Let f(x) be function inside ONE Integral on the RHS. Then let X= -x.
In the given integral
click on thanks button above please
what is antisymmetric function?
how do u know that G(X) = -G(-X)
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