Constant of Integration — Question 10

PDF ↗

Question 10

Suppose that ff is continuous on an interval II and that ∫f(x)dx=F(x)+C.\int f(x)\,dx = F(x)+C. Show that the family of curves y=F(x)+Cy=F(x)+C consists of vertical translations of a single curve.

Original worksheet page 1: question and worked solution for 7-9-010
Show solutionHide solution

Question 10 - Solution

Consider two members of the family of antiderivatives: y1=F(x)+C1,y2=F(x)+C2,y_1=F(x)+C_1, \qquad y_2=F(x)+C_2, where C1C_1 and C2C_2 are constants.

Subtract the equations: y2−y1=(F(x)+C2)−(F(x)+C1)=C2−C1.y_2-y_1=(F(x)+C_2)-(F(x)+C_1)=C_2-C_1.

The difference C2−C1C_2-C_1 is a constant independent of xx. Thus, for every value of xx, the vertical distance between the two curves is the same.

This means that changing the constant of integration does not alter the shape of the graph, but only shifts it upward or downward by a fixed amount.

Therefore, the family of functions y=F(x)+Cy=F(x)+C represents all vertical translations of a single curve.

Different values of C produce vertical shifts of the same graph\boxed{\text{Different values of }C\text{ produce vertical shifts of the same graph}}

Original worksheet page 2: question and worked solution for 7-9-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.