What is the area of the region under the curve y=1/x + 1/(x+1), beween the lines x=1 and x=2 and x axis?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The area of the region enclosed by the curve y = 1/x + 1/(x+1), the lines x=1 and x=2 and the x-axis is the definite integral of y = 1/x + 1/(x + 1) between the limits x = 1 and x = 2.
Int[ y dx], x = 1 to x = 2
=> Int [ 1/x + 1/(x + 1) dx], x = 1 to x = 2
=> ln|x| + ln|x + 1|, x = 1 to x = 2
=> ln 2 + ln 3 - ln 1 - ln 2
=> ln 3 - ln 1
=> ln 3
The required area is ln 3
giorgiana1976 | College Teacher | (Level 3) Valedictorian
In other words, we'll have to determine the area of the region bounded by the given curve, x axis and the lines x = 1 and x =2.
We'll evaluate the definite integral of the function y between the limits of integration: x = 1 to x = 2.
Int ydx = Int [1/x + 1/(x+1)]dx
We'll apply the additive property of integrals:
Int [1/x + 1/(x+1)]dx = Int dx/x + Int dx/(x+1)
Int dx/x + Int dx/(x+1) = ln |x| + ln|x + 1|
We'll apply Leibniz Newton formula to evaluate the definite integral:
Int dx/x + Int dx/(x+1) = F(2) - F(1)
Int dx/x + Int dx/(x+1) = ln 2 + ln 3 - ln 1 - ln 2
But ln 1 = 0
We'll reduce like terms and we'll get:
Int dx/x + Int dx/(x+1) = ln 3
The area of the region bounded by the given curve, x axis and the lines x = 1 and x =2 is A = ln 3 square units.