Use the fundamental theorem of calculus to find the area of the region bounded by the x-axis and the graph of y=4x^3-4x
Answer =
we need the x-intercepts.
4x^3 - 4x = 0
4x(x^2 - 1) 0
4x(x+1)(x-1) = 0
x = 0, -1, +1
If you sketch the curve, you will see two identical loops of the curve
cutting the x-axis, one above and one below.
The area of the loop at the left is
∫ (4x^3 - 4x) dx from -1 to 0
= [x^4 - 2x^2] from -1 to 0
= 0 - (1 - 2) = 1
so the area of the region bounded by the curve and the x-axis
= 2(1)
= 2 units^2
To find the area of the region bounded by the x-axis and the graph of y = 4x^3 - 4x, we can use the fundamental theorem of calculus. This theorem provides a relationship between the integral and the antiderivative of a function.
First, let's find the antiderivative of the function y = 4x^3 - 4x. To do this, we integrate each term separately:
∫(4x^3 - 4x) dx = ∫4x^3 dx - ∫4x dx
The antiderivative of 4x^3 is (4/4)x^4 = x^4, and the antiderivative of 4x is (4/2)x^2 = 2x^2. Therefore,
∫(4x^3 - 4x) dx = x^4 - 2x^2 + C,
where C represents the constant of integration.
Now that we have the antiderivative, we can find the area using the fundamental theorem of calculus. The area between two points, a and b, on the x-axis is given by the definite integral:
Area = ∫[a, b] (4x^3 - 4x) dx
Plugging in the antiderivative:
Area = [x^4 - 2x^2] [a, b]
Evaluating the antiderivative at the upper and lower limits:
Area = [b^4 - 2b^2] - [a^4 - 2a^2],
where [a, b] represents the evaluation of the expression at b and a respectively.
To find the area bounded by the x-axis and the graph of y = 4x^3 - 4x, you'll need to substitute the appropriate values of a and b into the equation and evaluate the expression.
To find the area of the region bounded by the x-axis and the graph of y=4x^3-4x, we can use the fundamental theorem of calculus.
Step 1: Find the antiderivative of the function y=4x^3-4x.
Taking the antiderivative of each term separately, we get:
∫(4x^3-4x)dx = ∫4x^3dx - ∫4xdx
Integrating each term:
∫4x^3dx = x^4 + C1 (where C1 is the constant of integration)
∫4xdx = 2x^2 + C2 (where C2 is the constant of integration)
Step 2: Subtract the antiderivative value at the lower limit from the antiderivative value at the upper limit.
To find the area between the x-axis and the graph of y=4x^3-4x, we need to evaluate the antiderivative at the upper and lower limits of integration.
Let's assume we're calculating the area between x = a and x = b.
According to the fundamental theorem of calculus, the area A is given by:
A = ∫[a, b](4x^3-4x)dx = [x^4 + C1] - [2x^2 + C2] evaluated from x = a to x = b.
So, the area of the region bounded by the x-axis and the graph of y=4x^3-4x is:
A = [b^4 + C1] - [2b^2 + C2] - ([a^4 + C1] - [2a^2 + C2])
Simplifying further:
A = b^4 - a^4 + (-2b^2 + 2a^2)
Thus, the answer is A = b^4 - a^4 - 2b^2 + 2a^2.