Hemolymph said:Homework Statement
from [-2,2] rad(4-x^2)
Homework Equations
The Attempt at a Solution
I know its a circle and i get the equation to be y^2+x^2=4
and I believe it has to be divided into a circle and rectangle
so the area of the rectangle i got to be 2
the circle i got to be 1/2(since its a half circle) times ∏(2)^2
the answer i got is
2+2pi (which is wrong) don't know where I went wrong tho.
Curious3141 said:Where did the rectangle come from? It's just a semicircular area.
The 1+ gave the rectangle there. You have no corresponding term here. Did you sketch the curve? Do you see a rectangle when you do?Hemolymph said:I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx
When you're using what you found in a "similar" problem, make sure it's actually similar to the one you're working on.Hemolymph said:I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx
And the solution had a the area broken up into a rectangle and a semicircle
Hemolymph said:I guess i tried to apply the same technique to this problem.
The answer came out to be 5+(25pi)/4
Evaluating an integral in terms of areas allows us to find the exact area between a curve and the x-axis, which can be useful in many real-world applications such as calculating volumes, work done, and average values.
To interpret an integral in terms of areas, we can visualize it as the sum of infinitely many rectangles with width dx and height f(x), where f(x) is the function being integrated. The area of each rectangle represents the value of the integral at that point, and the sum of all the rectangles gives us the total area under the curve.
Yes, for example, if we want to find the area under the curve y = x^2 from x = 1 to x = 3, we can evaluate the integral ∫x^2 dx by interpreting it in terms of areas. This would involve dividing the interval [1, 3] into smaller intervals and calculating the area of each rectangle, which would give us the total area under the curve.
Integrals and derivatives are inverse operations of each other. The derivative of a function gives us its rate of change, while the integral of a function gives us the total accumulation of that function over a given interval. The fundamental theorem of calculus states that these two operations are connected, where the derivative of the integral of a function is equal to the original function.
Yes, besides interpreting integrals in terms of areas, they can also be interpreted as the net change or accumulation of a quantity over an interval. This can be seen in applications such as finding the displacement of an object given its velocity function or calculating the total mass of an object given its density function.