Hi it's me again
I need help with this too and I promise to never use this site again! I feel so ashamed for asking :(
Use the Fundamental Theorem to evaluate the integral from 0 to 3 of the quantity x cubed minus 6 times x, dx.The answer has to have the antiderivative too.
Also Use a graph of the function to explain the geometric meaning of the value of the integral.
you need to find F(x) such that dF/dx = x^3-6x
Using the power rule, that is just
F(x) = 1/4 x^4 - 3x^2 + c
Now, the definite integral is just
F(3)-F(0) = (1/4 * 81 - 3*9 + c) - (0-0+c)
= 81/4 - 27
= -27/4
Read your text on the meaning of the integral. Consider that it is a sum of many many very thin rectangles, of width dx and height f(x).
Hello again! Please don't feel ashamed for asking questions – that's what I'm here for! I'm happy to help.
To evaluate the given integral using the Fundamental Theorem of Calculus, we first need to find the antiderivative of the given function. The antiderivative will represent the function whose derivative is equal to the integrand.
Let's find the antiderivative of the function f(x) = x^3 - 6x. To do this, we need to find a function F(x) such that F'(x) = f(x).
Step 1: Find the antiderivative of x^3.
The power rule tells us that if f(x) = x^n, where n ≠ -1, then the antiderivative F(x) is given by F(x) = (1/(n+1)) * x^(n+1). Applying this rule, the antiderivative of x^3 is F(x) = (1/(3+1)) * x^(3+1) = (1/4) * x^4.
Step 2: Find the antiderivative of -6x.
The power rule also applies here, so the antiderivative of -6x is F(x) = (-6/2) * x^2 = -3x^2.
Taking the sum of the antiderivatives, we have the antiderivative of f(x) = x^3 - 6x as F(x) = (1/4) * x^4 - 3x^2.
Now, using the Fundamental Theorem of Calculus, we can evaluate the definite integral from 0 to 3 of the function f(x) = x^3 - 6x:
∫[0 to 3] (x^3 - 6x) dx = F(3) - F(0) = [(1/4) * 3^4 - 3(3)^2] - [(1/4) * 0^4 - 3(0)^2]
= [(1/4) * 81 - 27] - [0 - 0]
= (81/4 - 27) - 0
= (81/4 - 108/4)
= -27/4
Therefore, the value of the given integral is -27/4, and the antiderivative is F(x) = (1/4) * x^4 - 3x^2.
Now, to explain the geometric meaning of the value of the integral, we can use a graph of the function f(x) = x^3 - 6x.
The integral of a function from a to b represents the area between the graph of the function and the x-axis over the interval [a, b]. In this case, the integral from 0 to 3 of f(x) represents the signed area between the graph of f(x) = x^3 - 6x and the x-axis over the interval [0, 3].
Since the value of the integral is negative (-27/4), it means that the signed area between the graph and the x-axis is below the x-axis. Visually, this means that the graph of f(x) dips below the x-axis and the area between the curve and the x-axis is negative.
I hope this explanation helps! Let me know if you have any further questions.