# If integral f =0, prove that f(x)=0 for all x in [a,b]

• Unassuming
In summary, if the integral of a continuous function f on [a,b] is equal to 0, then f(x) must equal 0 for all x in [a,b]. This can be proven by assuming the integral is 0 and using the definitions of lower and upper sums to show that they must also be equal to 0. However, this approach may be missing something and it may be better to use the Fundamental Theorem of Calculus, which has not yet been introduced. Alternatively, we can assume there is a point x0 in [a,b] where f(x0) is greater than 0 and use the epsilon-delta definition of continuity to construct a partition that shows the integral cannot be 0.
Unassuming

## Homework Statement

Suppose f is continuous on [a,b] with f(x) >= 0 for all x in [a,b].

If $$\int_a^b \! f = 0$$ , prove that f(x)=0 for all x in [a,b].

## The Attempt at a Solution

Assume $$\int_a^b \! f = 0$$.

Then the lower and upper sums equal 0 also. I looked at the definitions of these and came up with,

$$\sum_{i=1}^n ( inf\{ f(x) : x_{i-1} \leq x \leq x_i \} )( \bigtriangleup x_i )=0$$.

This looks obvious to me but that usually means I'm missing something because I don't know where to use that f is continuous.

HINT: It may be better to use the Fundamental Theorem of Calculus.

Hootenanny said:
HINT: It may be better to use the Fundamental Theorem of Calculus.

I don't "know" that yet. I am supposed to prove this without that.

Suppose there is an x0 in [a,b] such that f(x0)=c>0. Can you show there is a lower sum that is greater than 0?

Dick said:
Suppose there is an x0 in [a,b] such that f(x0)=c>0. Can you show there is a lower sum that is greater than 0?

Should I define a partition? If f(x_0)=c and c > 0 then the function must go up to atleast c, and since it is continuous it does not "jump" there.

I was thinking of making a small partition of a,x_0, and b. I then know what the sup or inf values of the two rectangles (speaking of the hieghts) will be greater than or equal to x_0 for the upper sum.

Would this do it? I figure that the upper sum and the lower are equal since the integral in the problem was defined as existing?

Unassuming said:
Should I define a partition? If f(x_0)=c and c > 0 then the function must go up to atleast c, and since it is continuous it does not "jump" there.

I was thinking of making a small partition of a,x_0, and b. I then know what the sup or inf values of the two rectangles (speaking of the hieghts) will be greater than or equal to x_0 for the upper sum.

Would this do it? I figure that the upper sum and the lower are equal since the integral in the problem was defined as existing?

No. Knowing a particular upper sum is greater than zero doesn't get
you anywhere all you know is that the integral is less or equal to
any upper sum.

But, yes. Make a partition. Use the epsilon-delta definition of continuity to show there is an interval around x0 such that f>c/2 on that interval. The use the endpoints of that interval to define the partition.

## 1. What does it mean for the integral of f to equal 0?

When the integral of a function f is equal to 0, it means that the area under the curve of f is equal to 0 over the given interval [a,b]. This can also be interpreted as the function f having a net balance of 0 over the interval, meaning that any positive values are balanced out by an equal amount of negative values.

## 2. How does this statement relate to the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus states that if a function f is continuous on a closed interval [a,b], and F is its antiderivative, then the definite integral of f from a to b is equal to F(b) - F(a). In this case, if the integral of f is equal to 0, then F(b) - F(a) = 0, meaning that F(b) = F(a). Since F is continuous, this implies that F(x) is constant over the interval [a,b]. And since F'(x) = f(x), this means that f(x) = 0 for all x in [a,b].

## 3. Can you provide an example to illustrate this statement?

One example is the function f(x) = sin(x) over the interval [0,π]. The integral of sin(x) from 0 to π is equal to 0, since the area above the x-axis is balanced out by the area below the x-axis. Therefore, by the statement, f(x) = sin(x) must also equal 0 for all x in [0,π].

## 4. Is this statement always true, or are there any exceptions?

This statement is always true as long as the function f is continuous over the given interval [a,b], and the integral is taken over the same interval. If either of these conditions are not met, the statement may not hold true.

## 5. How does this statement relate to the concept of a function's average value?

The average value of a function f over an interval [a,b] is defined as 1/(b-a) * the integral of f from a to b. If the integral of f is equal to 0, then the average value of f over [a,b] is also equal to 0. This can be interpreted as the function f having an equal balance of positive and negative values, resulting in an average of 0 over the interval.

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