Find the area bounded by the x-axis ,the curve y= 1/x+2, x=0,y=0, and x=5
To find the area bounded by the x-axis, curve y = 1/x+2, x = 0, y = 0, and x = 5, we need to integrate the function between the given limits of integration.
First, let's find the x-intercepts of the curve by setting y = 0:
0 = 1/x + 2
-2 = 1/x
x = -1/2
So, the curve intersects the x-axis at the point (-1/2, 0).
Next, we'll integrate the function y = 1/x+2 within the given limits [0, 5]. However, since the function y = 1/x+2 is not defined at x = 0, we need to split the integral into two parts: from 0 to -1/2, and from -1/2 to 5.
Integral from 0 to -1/2: ∫[0, -1/2] (1/x+2) dx
Integral from -1/2 to 5: ∫[-1/2, 5] (1/x+2) dx
To integrate the function, we can use the natural logarithm. The integral of 1/x is ln|x|, so we can rewrite the integrals as:
Integral from 0 to -1/2: ∫[0, -1/2] (1/x+2) dx = [ln|x+2|] from 0 to -1/2
Integral from -1/2 to 5: ∫[-1/2, 5] (1/x+2) dx = [ln|x+2|] from -1/2 to 5
Now, let's evaluate each part of the integral.
Integral from 0 to -1/2: [ln|-1/2 + 2|] - [ln|0 + 2|]
= [ln|(3/2)|] - [ln|2|]
= ln(3/2) - ln(2)
= ln(3/2) - ln(2)
Integral from -1/2 to 5: [ln|5 + 2|] - [ln|-1/2 + 2|]
= [ln|7|] - [ln|(3/2)|]
= ln(7) - ln(3/2)
= ln(7) - ln(3/2)
To find the area, we need to subtract the smaller integral value from the larger one:
Area = [ln(7) - ln(3/2)] - [ln(3/2) - ln(2)]
= ln(7) - ln(3/2) - ln(3/2) + ln(2)
= ln(7) - 2ln(3/2) + ln(2)
Thus, the area bounded by the x-axis, curve y = 1/x+2, x = 0, y = 0, and x = 5 is ln(7) - 2ln(3/2) + ln(2).