10). Find the area of the region bounded by the graph of f(x)=4lnx/x, y=0, x=5.8
11). Find the area of the region bounded by x=y^2-1y, x=0
To find the area of a region bounded by curves, we use definite integrals.
10) To find the area of the region bounded by the graph of f(x) = 4ln(x)/x, y = 0, and x = 5.8, we follow these steps:
Step 1: Identify the limits of integration. Since the region is bounded by x = 5.8, we need to find the y-values that correspond to the intersection points. To do this, we set the equation of the curve equal to y = 0 and solve for x:
4ln(x)/x = 0
ln(x)/x = 0
ln(x) = 0
x = 1
So, the integral will be taken from x = 1 to x = 5.8.
Step 2: Set up the integral. The area is given by the integral of the curve equation f(x) minus the lower boundary y = 0:
Area = ∫[1, 5.8] (f(x) - 0) dx
Step 3: Evaluate the integral. The integral of 4ln(x)/x with respect to x will require the use of integration techniques. After evaluating the integral, substitute the upper and lower limits:
Area = [∫(f(x) - 0) dx] [1, 5.8]
11) To find the area of the region bounded by the curves x = y^2 - 1 and x = 0, we follow these steps:
Step 1: Identify the limits of integration. Since the region is bounded by x = 0, we need to find the y-values that correspond to the intersection of the two curves. To do this, we set the equations of the curves equal to each other:
y^2 - 1 = 0
y^2 = 1
y = ±1
So, the integral will be taken from y = -1 to y = 1.
Step 2: Set up the integral. The area is given by the integral of the curve equation x minus the lower boundary x = 0:
Area = ∫[-1, 1] (x - 0) dy
Step 3: Convert the integral to be with respect to y. The equation x = y^2 - 1 is written in terms of x, so we need to rewrite the integral with respect to y:
Area = ∫[-1, 1] [(y^2 - 1) - 0] dy
Step 4: Evaluate the integral. After integrating and substituting the upper and lower limits, you can calculate the area:
Area = [∫[(y^2 - 1) - 0] dy] [-1, 1]