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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
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- Jan 31, 2012

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- Mar 1, 2012

- 655

hint ...

$\displaystyle \int_b^x g(t) \, dt = -\int_x^b g(t) \, dt$

$\displaystyle \int_b^x g(t) \, dt = -\int_x^b g(t) \, dt$

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- Jan 31, 2012

- 2,678

so then

$\displaystyle \int_b^x g(t) \, dt +\int_x^b g(t) \, dt=0$ |

- Mar 1, 2012

- 655

that’s not the problem

- Mar 1, 2012

- 655

- Mar 1, 2012

- 655

Properties of definite integrals ...