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If this is on mymaths.co.uk then the answer to this question is: Integration.

That is how to find the area under the curve.

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14y ago

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What I helps you to find the area under the curve y x3 4x - 3.6?

"integration"


What I helps you find the area under the curve y x3 4x-3.6?

Integration.


What i helps you to find the area under the curve yx3 4x-3.6?

if its the mymaths one type integration into the box!


What is the i that helps you find the area of the curve y?

integral.


What I helps you to find the area under the curve y x3 4x-3.6?

if this is for www.mymaths.co.uk, integration is all you need. best regards, lokilotus


What helps you to find the area under a curve yx34x-3.6?

37.6


What is the explanation for the area under the curve?

In statistics you can find the area under a curve to establish what to expect between two input numbers. If there is a lot of area under the curve the graph is tall and there is a higher probability of things occurring there than when the graph is low.


Find the area under the standard normal curve between -1.33 and the mean P(-1.33?

To find the area under the standard normal curve between -1.33 and the mean (0), we can use the standard normal distribution table or a calculator. The area to the left of -1.33 is approximately 0.0918. Since the total area under the curve is 1 and the curve is symmetrical around the mean, the area between -1.33 and 0 is about 0.5 - 0.0918 = 0.4082. Thus, the area under the curve from -1.33 to the mean is approximately 0.4082.


What I helps you to find the area under the curve y x3 4x 3.6?

well, it would help to clarify your question, but integrate the fuction and solve on the interval given


X equals acos3m y equals bsin3m find area under curve?

(3ab*pi)


Find position given velocity vs time graph?

To find the position from a velocity-vs-time graph, you need to calculate the area under the velocity curve. If the velocity is constant, the position can be found by multiplying the velocity by the time. If the velocity is changing, you need to calculate the area under the curve using calculus to determine the position.


Find the area under a normal distribution curve that lies within one standard deviation of the mean is approximattely?

The area under N(0,1) from -1 to 1 = 0.6826

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