Does anyone know how to do this question? If so please show me how, I have a test tomorrow :/
Calculate the total are between the graph of y=-x^3 and the x-axes over (-1,2)
Calculate the total are between the graph of y=-x^3 and the x-axes over (-1,2)
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I'm assuming since you said total area that it wants all the area bounded by the x axis and the function y=-x^3, regardless if the area is technically negative. So in this problem I will assume area is always positive. At the end, however, I will give the net area in case that is what it meant.
You should start out by graphing y=-x^3 with 4 points, namely (-1,1), (0,0), (1,-1), and (2,-8)
You'll notice that if you draw vertical lines at x=-1 and x=2 you can see the area bounded between the function and the x axis. You'll also notice that some of this area is above the x axis and some is below. So the positive area is on the interval [-1,0] and the negative area is on the interval [0,2].
Recall that integrating a function from a certain x value to another x value gives the area under the graph.
So to find the area on the interval [-1,0] you need to integrate from -1 to 0.
Int(-x^3)dx = -x^4/4. Evaluate this from -1 to 0 you get [0-(-1/4)] = 1/4. So this is the area under the graph on the first interval [-1,0].
Now we need to find the area between the function and x axis on the interval [0,2].
You'll notice that this area is below the x axis so it will be negative. So we need the absolute value of this area. So we find the absolute value of the integral Int(-x^3)dx from 0 to 2 and add this answer to 1/4.
Int(-x^3)dx = -x^4/4. Evaluated from 0 to 2 equals...[-16/4-0]= -4.
And the absolute value to -4 is 4.
4+ 1/4 = 16/4 + 1/4 = 17/4.
So 17/4 is the total area bounded between the function and x axis.
However, if it was asking for the net area then you can see from the graph that it has more negative area than positive area so we should expect a negative net area. As we saw, the area on the interval [-1,0] was positive and was 1/4. But the area on the interval [0,2] was technically -4. So the net area is -4+1/4 = -16/4 + 1/4 = -15/4.
So -15/4 is the net area and 17/4 is the real-life practical area.
Hope that helped.
You should start out by graphing y=-x^3 with 4 points, namely (-1,1), (0,0), (1,-1), and (2,-8)
You'll notice that if you draw vertical lines at x=-1 and x=2 you can see the area bounded between the function and the x axis. You'll also notice that some of this area is above the x axis and some is below. So the positive area is on the interval [-1,0] and the negative area is on the interval [0,2].
Recall that integrating a function from a certain x value to another x value gives the area under the graph.
So to find the area on the interval [-1,0] you need to integrate from -1 to 0.
Int(-x^3)dx = -x^4/4. Evaluate this from -1 to 0 you get [0-(-1/4)] = 1/4. So this is the area under the graph on the first interval [-1,0].
Now we need to find the area between the function and x axis on the interval [0,2].
You'll notice that this area is below the x axis so it will be negative. So we need the absolute value of this area. So we find the absolute value of the integral Int(-x^3)dx from 0 to 2 and add this answer to 1/4.
Int(-x^3)dx = -x^4/4. Evaluated from 0 to 2 equals...[-16/4-0]= -4.
And the absolute value to -4 is 4.
4+ 1/4 = 16/4 + 1/4 = 17/4.
So 17/4 is the total area bounded between the function and x axis.
However, if it was asking for the net area then you can see from the graph that it has more negative area than positive area so we should expect a negative net area. As we saw, the area on the interval [-1,0] was positive and was 1/4. But the area on the interval [0,2] was technically -4. So the net area is -4+1/4 = -16/4 + 1/4 = -15/4.
So -15/4 is the net area and 17/4 is the real-life practical area.
Hope that helped.