find the area enclosed between the following curves:- Y= lnx and y= -2x+3 and the ordinate x=3
The graphs intersect at x = 1.35, so we want
Int(ln x - (3-2x))[1.35,3]
= (xlnx - x - 3x + x^2)[1.35,3]
= xlnx - 4x + x^2 [1.35,3]
= 0.296 - (-3.172) = 3.468
hi, thanks for the answer but the question asks for it to be solved in integration form- is this right?
To find the area enclosed between the two curves, we need to first determine the points of intersection between the curves.
Setting the two equations equal to each other gives us:
lnx = -2x + 3
To solve for x, we can start by using the property of logarithms to rewrite lnx as e^(lnx), where e is the natural logarithm base:
e^(lnx) = e^(-2x + 3)
This simplifies to:
x = e^(-2x + 3)
Now, we can use a numerical method like Newton's method, bisection, or a graphing calculator to find the approximate value of x where the curves intersect. Let's assume the point of intersection is x = a.
Next, we need to determine the interval of integration. Since we know one of the boundaries is x = 3, we need to find the other boundary. To do this, we check if the curves intersect at x = 3. If they do, then the interval of integration is from the x-value of the intersection point we found earlier (x = a) to x = 3. If they don't intersect at x = 3, then the interval is from the minimum x-value of the two curves to x = 3.
Once we have the interval of integration, we can set up the definite integral to find the area enclosed between the curves:
Area = ∫[a, 3] (f(x) - g(x)) dx
Where f(x) is the upper curve (in this case, y = -2x + 3) and g(x) is the lower curve (y = lnx).
Evaluating this integral will give us the area enclosed between the curves.