Asked by anonymous
How do you find the area enclosed by the y axis line, y= the square root of x+2, y=1/(x+1), and x=2?
Answers
Answered by
anonymous
would I just use the too functions for the top and bottom function for area and then use the 2 and 0 as limits?
Answered by
oobleck
yes
∫[0,2] √(x+2) - 1/(x+1) dx
∫[0,2] √(x+2) - 1/(x+1) dx
Answered by
mathhelper
anonymous said:
area = ∫ (x+2)^(1/2) - 1/(x+1) from 0 to 2
= [(2/3)(x+2)^(3/2) - ln(x+1) ] from 0 to 2
= (2/3)(8) - ln(3) - ( (2/3)2√2 - ln1)
= 16/3 - ln(3) - 2/3√2 - 0
= 16/3 - ln3 - 2√2/3
= appr 2.3491
confirmed by Wolfram:
www.wolframalpha.com/input?i=+%E2%88%AB+%28x%2B2%29%5E%281%2F2%29++-+1%2F%28x%2B1%29dx+from+0+to+2
area = ∫ (x+2)^(1/2) - 1/(x+1) from 0 to 2
= [(2/3)(x+2)^(3/2) - ln(x+1) ] from 0 to 2
= (2/3)(8) - ln(3) - ( (2/3)2√2 - ln1)
= 16/3 - ln(3) - 2/3√2 - 0
= 16/3 - ln3 - 2√2/3
= appr 2.3491
confirmed by Wolfram:
www.wolframalpha.com/input?i=+%E2%88%AB+%28x%2B2%29%5E%281%2F2%29++-+1%2F%28x%2B1%29dx+from+0+to+2
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.