# Find the constant of integration #c# given that #f''(x)=2x# and the points #(1,0)# and #(0,5)# lie on the curve ?

# f(x) = 1/3x^3 -16/3x + 5 #

We have:

Given we have a second derivative (we essentially have a Second Order twice separable Ordinary Differential Equation) we can integrate twice which will introduce two constants of integration. As we have two coordinates that the curve passes through we can then we can find both constants and form a full solution.

Hence we have the complete solution:

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To find the constant of integration, integrate the given function twice and use the provided points to solve for the constant.

The antiderivative of f''(x) = 2x is f'(x) = x^2 + C1, where C1 is a constant of integration.

Integrating again, we get f(x) = (1/3)x^3 + C1x + C2, where C2 is another constant of integration.

Given the points (1,0) and (0,5), we can substitute these into the equation.

For (1,0): 0 = (1/3)1^3 + C1(1) + C2 0 = 1/3 + C1 + C2 0 = 1/3 + C1 + C2 (1)

For (0,5): 5 = (1/3)0^3 + C1(0) + C2 5 = 0 + 0 + C2 5 = C2

Substitute C2 = 5 into equation (1) to solve for C1: 0 = 1/3 + C1 + 5 -5 1/3 = C1 C1 = -16/3

Therefore, the constant of integration is C1 = -16/3.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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