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?
3 answers
would I just use the too functions for the top and bottom function for area and then use the 2 and 0 as limits?
yes
∫[0,2] √(x+2) - 1/(x+1) dx
∫[0,2] √(x+2) - 1/(x+1) dx
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