Q: Prove $\int_{x}^{x+h} f(t)dt= \int_a^b f(t)dt$ using only $\int_a^b f(t)dt=0$. Prove the following theorem: $\int_{x}^{x+h} f(t)dt= \int_a^b f(t)dt$, if $\int_a^b f(t)dt=0$ and $h \in R$. My idea was to use induction. Let $k=1$. Then, by hypothesis, $\int_a^b f(t)dt=0$. So $\int_a^{a+h} f(t)dt=0$ which can be written as $$h\int_a^{a+h}f(t)dt=0$$ So $\int_a^{a+h}f(t)dt=0$. My intuition tells me I need to use the fact that $f$ is continuous on $a,b$ and then use that fact that $$F(x+a)-F(x)= \lim_{\delta \to 0} \int_x^{x+\delta} f(t)dt$$ A: One does not need continuity. Simply apply the definition of definite integral $$\int_x^{x+h} f(t)\,dt=F(x+h)-F(x)$$ where $F$ is antiderivative of $f$, so that $F'(t)=f(t)$ and $$\int_a^b f(t)\,dt=\int_a^b F'(t)\,dt=\left.F(t)\right|_a^b=F(b)-F(a)=0$$ A: In general, one does not have that $F(x + h) - F(x) = \int_x^{x + h} f(t)\,dt$. For a counterexample, consider $f(x) = \sin x$. In this case, $F(x) = \frac12 x^2$. The standard definition of definite integral is $$\int_a^b f(t)\,dt = \lim_{P\to\infty}\sum_{k=1}^n f(x_k)(x_{k+1}-x_k)$$ where $a = x_1 < x_2 < \cdots < x_n = b$ and $\{x_1,\dots,x_{n-1}\}$ is a partition of $[a,b]$. However, with the condition that $f$ has zero integral over $[a,b]$, we can deduce the following. $$\int_a^b f(t)\,dt = \int_{a}^b f(t)\,dt = f(x) (b-a) + \int_a^b f(t)\,dt = 0(b-a) = 0.$$ By convention, an integral over an interval of the form $[a,\infty)$ (or $[-\infty,a)$) is $+\infty$ if $f$ is nonzero there, and $0$ if $f$ is zero there. In any case, it still equals to $0$. Here is an example for a nonnegative, continuous function which has a positive integral over some interval. Define $$f(x) = \begin{cases} 1 & x\ge0,\\ -1 & x<0. \end{cases}$$ Then $\int_0^1 f(t)\,dt = \int_0^1 \frac{1}{x} \,dx = \infty$. A: As others have mentioned, the integral is well-defined only if the integral over some intervals of discontinuity vanishes. But if $f$ is continuous and nonnegative on $[a,b]$, this condition is equivalent to $f(x)>0$ on $[a,b]$: If $f$ is not identically $0$, $f(x)$ must have a positive sign somewhere, and hence can't be zero at all. So if $f$ is continuous, $f(x)\geq 0$ on $[a,b]$ and $f(x)\not=0$ for any $x\in[a,b]$, then $$\int_a^b f(x)\,dx>0$$ However, if $f(x)\geq 0$ on $[a,b]$ and $f(x)=0$ somewhere in $(a,b)$, then the value of the integral can indeed be $0$. Take, for instance, $$f(x)=\frac{\sin x}{x}$$ Then $f$ is continuous on $[0,\infty)$ and $f(x)\geq 0$ on $[0,\infty)$ and it has a $0$ where it crosses the discontinuity at $x=0$. Hence $$\int_0^\infty f(x)\,dx=0$$ For this example, this is exactly the reason that the integral "fails" to converge because it oscillates infinitely many times. So the condition that the integral of $f$ over $[a,b]$ is zero in the statement of the theorem is equivalent to $f$ being identically zero, hence $F$ is identically $0$ so the integral over any interval is $0$. Hence the integral of $f$ over $[a,b]$ can indeed be zero. However, note that this condition is not enough for an integral of $f$ to be defined: it's still not clear that the function obtained by integrating $f$ over $[a,b]$ is differentiable on $(a,b)$. To answer one of the commentors who asked for an example of a function for which $F$ is not differentiable at $0$ but the integral of $f$ over any interval is $0$, try the function $$f(x)=\left\{\begin{array}{cc} \sin \frac{1}{x} & x\neq 0\\ 0 & x=0 \end{array}\right.$$ then $\int_0^\infty f(x)\,dx=0$, but since $F(x)=\int_0^x f(t)\,dt$ is not differentiable at $0$, there is no guarantee that $F$ will be differentiable everywhere. P.S. This means that the condition that the integral of $f$ is zero for all intervals, which is required in the statement of the theorem, is a sufficient condition for an integral to exist, but not a necessary condition. Since the statement of the theorem says "if $f$ has zero integral over $[a,b]$, then...", it looks like it is a necessary condition. But it isn't. For instance, if the definition of the Riemann integral is instead extended to include any function which is integrable over every interval, then the function $F$ given above is integrable on $[a,b]$ and its integral over $[a,b]$ is zero.