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Intergal question
- Thread starter ringyal1
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ringyal1 said:If f is a continuous function, show that definite integral 0to 1 of f(x)dx = integral 0 to 1 of f(1-x)dx
Kindly help me solve this problems thanks...
Are you sure you have the limits correct?
BigGlenntheHeavy
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\(\displaystyle Prove \ \int_{0}^{1}f(x)dx \ = \ \int_{0}^{1}f(1-x)dx\)
\(\displaystyle \int_{0}^{1}f(x)dx \ = \ F(1)-F(0), \ FTC \ I\)
\(\displaystyle \int_{0}^{1}f(1-x)dx, \ let \ u \ = \ 1-x, \ then \ du \ = \ -dx \ \implies \ -du \ = \ dx\)
\(\displaystyle \int_{0}^{1}f(1-x)dx \ = \ -\int_{1}^{0}f(u)du \ = \ \int_{0}^{1}f(u)du \ = \ F(1)-F(0)\)
\(\displaystyle Hence, \ \int_{0}^{1}f(x)dx \ = \ \int_{0}^{1}f(1-x)dx, \ QED\)
\(\displaystyle \int_{0}^{1}f(x)dx \ = \ F(1)-F(0), \ FTC \ I\)
\(\displaystyle \int_{0}^{1}f(1-x)dx, \ let \ u \ = \ 1-x, \ then \ du \ = \ -dx \ \implies \ -du \ = \ dx\)
\(\displaystyle \int_{0}^{1}f(1-x)dx \ = \ -\int_{1}^{0}f(u)du \ = \ \int_{0}^{1}f(u)du \ = \ F(1)-F(0)\)
\(\displaystyle Hence, \ \int_{0}^{1}f(x)dx \ = \ \int_{0}^{1}f(1-x)dx, \ QED\)
BigGlenntheHeavy
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Let u =1-x, then where ever x = zero, u =1 and where ever x =1, u =0.