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So, the easiest way to start thinking about this is by looking at straight lines. The gradient of a straight line is just its slope, which is the difference in the y values between any two points...
For the x2 term we add one to the power, and then divide by 3 to get x3/3 For the next term we add one to the power and divide by the new power to get 3x2/2 We do the same...
To find the gradient of a curve, you simply differentiate the equation of the curve. The first thing I like to do in any differentiation question is to simplify each expression where you can i.e. whenever...
Integrate to get y(x) = (1/3)x^3 -2x+c where c is a constant. Substitute in our data 7 =y(0) = (1/3)(0)^3 -2*(0) +c = c. So y(x) =(1/3)x^3 -2x+7 and therefore y(3) = (1/3)(3)^3 -2*3 +7 = 9-6+7 = 10
To find the area under a curve that is bounded by the x-axis you simply need to integrate the equation of the curve between the limits, so for this equation we will integrate y=x3-3x2Answered by Jack T. • Maths tutor12930 Views
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