True or false:
If f(x) is continuous on the interval [a,b] & if the definite integral of f(x) over this interval has value zero, then the equation f(x)=0 has at least one solution in the interval [a,b].
True or false:
If f(x) is continuous everywhere and F(x)= int from 0 to x of f(t)dt then the equation F(x)=0 has at least one solution.
True for both statements.
For the first statement, we can use the Intermediate Value Theorem to show that if f(x) is continuous on the interval [a, b] and the definite integral of f(x) over this interval is zero, then the equation f(x) = 0 must have at least one solution in the interval [a, b]. The Intermediate Value Theorem states that if a continuous function takes on two different values at two points, then it must also take on every value between those two points. Since the integral of f(x) over [a, b] is zero, it means that the function crosses the x-axis (f(x) = 0) somewhere between a and b.
To prove this, we can use a proof by contradiction. Assume that f(x) ≠ 0 for all x in [a,b]. Since f(x) is continuous, it either remains positive or negative throughout the interval. Without loss of generality, let's assume it remains positive. In this case, the integral of f(x) over [a, b] will be positive as well (since it is the area under the curve). But this contradicts the assumption that the integral is zero. Therefore, our assumption that f(x) ≠ 0 for all x in [a,b] must be false, and f(x) = 0 has at least one solution in the interval [a, b].
For the second statement, we can also use the Intermediate Value Theorem. Since f(x) is continuous everywhere, we can define F(x) as the integral of f(t) from 0 to x. If F(x) = 0 for some x, it means that the definite integral of f(t) from 0 to x is zero. Using the same reasoning as in the first statement, we can conclude that since F(x) = 0, the equation f(t) = 0 must have at least one solution between 0 and x. Thus, the equation F(x) = 0 has at least one solution.