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

Elijah Abram · Accepted Answer

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

We have: # f''(x) = 2x # ..... [A] 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. Integrate [A] wrt #x# to gain: # f'(x) = x^2 + A # Integrate this result wrt #x# to gain: # f(x) = x^3/3 + Ax + B # Using the first coordinate #(1,0)#, we know that: # f(1) = 0 => (1)^3/3 + A(1) + B = 0 # # :. 1/3+A+B = 0# Using the second coordinate #(0,5)#, we know that: # f(0) = 5 => (0)^3/3 + A(0) + B = 5 # # :. B = 5# Substituting #B=5# into the first equation we get: # :. 1/3+A+5 = 0 => A = -16/3# Hence we have the complete solution: # f(x) = 1/3x^3 -16/3x + 5 #

Isabella Ackerman · Answer

undefined 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.

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