Seed5813
New member
- Joined
- Jan 29, 2018
- Messages
- 24
Ok so I evaluated the indefinite integral and got the right answer but when I plugged in the bounds, I got the wrong answer.
Calculate the definite integral:
. . . . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, =\, F(x)\, =\, \int_1^3\, \dfrac{x\, +\, 1}{x\, (x^2\, +\, 1)}\, dx\)
Computed by Maxima:
. . .\(\displaystyle \displaystyle \int\, f(x)\, dx\, \)
. . . . . . .\(\displaystyle =\, \ln\left(\big|x\big|\right)\, -\, \dfrac{\ln(x^2\, +\, 1)}{2}\, +\, \arctan(x)\, +\, C\)
. . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, \)
. . . . . . .\(\displaystyle =\, \ln(3)\, -\, \dfrac{\ln(10)}{2}\, +\, \arctan(3)\, +\, \dfrac{2\ln(2)\, -\, \pi}{4}\)
. . . . . . .\(\displaystyle \approx 0.7575409414518656\)
For the last term I had arctan(1) which is exactly double the last term the int calculator had.
Sent from my LGLS755 using Tapatalk
Calculate the definite integral:
. . . . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, =\, F(x)\, =\, \int_1^3\, \dfrac{x\, +\, 1}{x\, (x^2\, +\, 1)}\, dx\)
Computed by Maxima:
. . .\(\displaystyle \displaystyle \int\, f(x)\, dx\, \)
. . . . . . .\(\displaystyle =\, \ln\left(\big|x\big|\right)\, -\, \dfrac{\ln(x^2\, +\, 1)}{2}\, +\, \arctan(x)\, +\, C\)
. . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, \)
. . . . . . .\(\displaystyle =\, \ln(3)\, -\, \dfrac{\ln(10)}{2}\, +\, \arctan(3)\, +\, \dfrac{2\ln(2)\, -\, \pi}{4}\)
. . . . . . .\(\displaystyle \approx 0.7575409414518656\)
For the last term I had arctan(1) which is exactly double the last term the int calculator had.
Sent from my LGLS755 using Tapatalk
Last edited by a moderator: