Find the area of the region bounded by the graph of f(x)=x(x+3)(x+1) and the x-axis on the interval [-3,0]
To find the area of the region bounded by the graph of f(x)=x(x+3)(x+1) and the x-axis on the interval [-3,0], we need to integrate the function from -3 to 0.
Step 1: Find the integral of the function f(x).
The integral of f(x) can be found using the power rule for integration:
∫(x(x+3)(x+1)) dx = ∫(x^3 + 4x^2 + 3x) dx
Step 2: Integrate each term separately.
∫x^3 dx = (1/4)x^4
∫4x^2 dx = (4/3)x^3
∫3x dx = (3/2)x^2
Step 3: Evaluate the definite integral from -3 to 0.
To find the area, we substitute the limits of integration into the antiderivative we found in step 2 and subtract the two results:
Area = ∫[0 to -3] (x^3 + 4x^2 + 3x) dx = [((1/4)x^4 + (4/3)x^3 + (3/2)x^2)] [from 0 to -3]
Area = ((1/4)(-3)^4 + (4/3)(-3)^3 + (3/2)(-3)^2) - ((1/4)(0)^4 + (4/3)(0)^3 + (3/2)(0)^2)
Step 4: Simplify the expression.
Area = ((81/4) + (36/3) + (27/2)) - (0 + 0 + 0)
Area = (81/4 + 48/4 + 27/2)
Area = (81+48+54)/4
Area = 183/4
Therefore, the area of the region bounded by the graph of f(x)=x(x+3)(x+1) and the x-axis on the interval [-3,0] is 183/4.
To find the area of the region bounded by the graph of the function f(x) = x(x+3)(x+1) and the x-axis on the interval [-3,0], we can use the definite integral. The definite integral of a function represents the signed area between the function and the x-axis over a given interval.
First, we need to determine the points where the function intersects the x-axis. For f(x) = x(x+3)(x+1) to be equal to zero, either x, (x+3), or (x+1) must be zero.
Setting x = 0, we get f(0) = 0(0+3)(0+1) = 0. So, the point (0,0) is one of the x-intercepts.
Setting x + 3 = 0, we find x = -3. Thus, the point (-3,0) is another x-intercept.
Setting x + 1 = 0, we get x = -1. Therefore, the point (-1,0) is another x-intercept.
Now, let's integrate the absolute value of the function f(x) = |x(x+3)(x+1)| from x = -3 to x = 0, as the absolute value ensures that the area is positive.
The definite integral of f(x) = |x(x+3)(x+1)| from x = -3 to x = 0 can be written as:
∫[-3,0] |x(x+3)(x+1)| dx
To evaluate this integral, we need to break it up into separate intervals based on the x-intercepts.
1. From x = -3 to x = -1:
∫[-3,-1] |x(x+3)(x+1)| dx
2. From x = -1 to x = 0:
∫[-1,0] |x(x+3)(x+1)| dx
Since the function is symmetric, we can integrate in one section and double the result to get the total area:
2 * ∫[-3,-1] |x(x+3)(x+1)| dx
To evaluate the integral, we can simplify the function inside the absolute value brackets:
f(x) = x(x+3)(x+1)
Now we have the function that we can integrate from x = -3 to x = -1.
Finally, we use a numerical method (such as Riemann sums) or a computer program to integrate the function and find the area of the region bounded by the graph of f(x) and the x-axis on the interval [-3,0].
just expand the polynomial and integrate
f(x) = x^3+4x^2+3x