Constant of Integration — Question 9

PDF ↗

Question 9

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 if two antiderivatives of ff agree at a single point, then they are identical on the entire interval.

Original worksheet page 1: question and worked solution for 7-9-009
Show solutionHide solution

Question 9 - Solution

Let FF and GG be two antiderivatives of ff on the interval II. Then F′(x)=f(x)andG′(x)=f(x)for all x∈I.F'(x)=f(x) \quad\text{and}\quad G'(x)=f(x) \quad\text{for all }x\in I.

Assume that there exists a point x0∈Ix_0\in I such that F(x0)=G(x0).F(x_0)=G(x_0).

Consider the function H(x)=F(x)−G(x).H(x)=F(x)-G(x).

Differentiate H(x)H(x): H′(x)=F′(x)−G′(x)=f(x)−f(x)=0for all x∈I.H'(x)=F'(x)-G'(x)=f(x)-f(x)=0 \quad\text{for all }x\in I.

Thus, HH has zero derivative on II, so HH is constant on II. That is, H(x)=Kfor all x∈I.H(x)=K \quad\text{for all }x\in I.

Evaluate HH at x0x_0: H(x0)=F(x0)−G(x0)=0.H(x_0)=F(x_0)-G(x_0)=0.

Hence, K=0K=0, and therefore F(x)−G(x)=0for all x∈I.F(x)-G(x)=0 \quad\text{for all }x\in I.

This implies F(x)=G(x)for all x∈I.F(x)=G(x) \quad\text{for all }x\in I.

F(x)≡G(x)\boxed{F(x)\equiv G(x)}

Original worksheet page 2: question and worked solution for 7-9-009

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