Cannot understand intuition why integral is the difference of two functions on both ends I cannot understand how it happens that the area under the curve of the function $f$ between $x=a$ and $x=b$ equals $f(b) - f(a)$. We simply evaluate functions on both ends and find the difference. How can it give us the area under the curve at this region?
I could not find any reasonable explanation. I just need the intuition to understand what is going on.
Thank you
 A: How much did you spend in March? Look at your bank balance at the end of the month, and at the beginning. Then subtract. Exact same thing. Although if your expenditure is $f(t)$ you should use a different name, say $F(t)$ for the balance.And you need a convention that expenditure can be negative (saving). But hopefully this analogy makes it seem a bit more intuitive.
A: The area under the graph of the function $f$ on the interval $[a,b]$ is defined as 
$$\int_a^bf(x)\,\mathrm{d}x$$
and if there exists a differentiable function $F$ such that $F'=f$, then this integral is the difference $F(b)-F(a)$.
Here the function $F$ is a totally different function which depends on the behavior of $f$.
A: Consider the function $f(x) = x$. We can use basic geometry to find the area under this line from $x=a$ to $x=b$: subtract a small triangle (with base and height $a$) from a large triangle (with base and height $b$). This gives
$$A = \frac12b^2 - \frac12a^2.$$
Observe that another way to write this would be
$$A = F(b) - F(a)$$
where $F(x) = \frac12x^2$, which is the antiderivative of $f(x)$.
Most functions will be more complicated than this, of course, so we typically can't do the calculation using basic geometry. But the principle is the same, and the fact that we can find the antiderivative using calculus is what makes this concept very powerful.
A: 
I've adapted Wikipedia's explanation, more general proof which doesn't require $y$ to be continuous but a bit complex also on Wikipedia
For any continuous curve $y=f(x)$ we can define the area function $A(x)$ to give area under the curve $y$ between $0$ and $x$ ($0$ is taken to fit the illustration but we can as well take any number instead of $0$).
For us to get the area beneath the curve from $x$ to $x+h$ we take the area from $0$ to $x+h$ and subtract it with the area from $0$ to $x$ so we get $A(x+h)-A(x)$.
Now the red area is approximated by the rectangle with height $f(x)$ and width $h$ so $A(x+h)-A(x)\approx f(x)\cdot h$ and we can find the exact area by adding the red excess.
$$A(x+h)-A(x)=f(x)\cdot h+\color{red}{\text{Red Excess}}$$
Now this can differently be written as
$$f(x)=\frac{A(x+h)-A(x)}{h}-\frac{\color{red}{\text{Red Excess}}}{h}$$
The red excess is smaller than the whole black rectangle (Excess) with sides $h$ and $f(x+h)-f(x)$ so we can write $$\left|f(x)-\frac{A(x+h)-A(x)}{h}\right|=\frac{|\color{red}{\text{Red Excess}}|}{h}\leq\frac{\text{Excess}}{h}=\frac{h|f(x+h)-f(x)|}{h}=\\|f(x+h)-f(x)|$$
By continuity we know $\lim_{h\to 0}|f(x+h)-f(x)|=0$  so we have that
$$\lim_{h\to 0}\left|f(x)-\frac{A(x+h)-A(x)}{h}\right|=0\\f(x)=\lim_{h\to 0}\frac{A(x+h)-A(x)}{h}$$
The right side is the definition of the derivative so $f(x)dx=A'(x)$ so we have that $\int f(x)=A(x)$. By definition the area between $a,b$ is $A(b)-A(a)$ so we have that the area is $\int_a^bf(x)dx=A(b)-A(a)$.
A: Area of a plane bounded region is a concept which is reasonably tough to define. The simpler way to define area is via the concept of Jordan content which is related to Riemann integral (the slightly difficult way is based on concept of Lebesgue measure which is related to Lebesgue integral). If you want to avoid these definitions you may straightaway jump to the part after last fold. 

Before we discuss the area under graph of a function it makes sense to define the area of plane regions and I will give here an exposition based on Jordan content. Suppose that $A$ is a set of points in plane which is bounded meaning that there exists real numbers $a, b, c, d$ with $a<b, c<d$ such that $$A\subseteq [a, b] \times [c, d] =\{(x, y) \mid x\in [a, b], y\in[c, d] \} $$ Next consider a partition $$P=\{x_0,x_1,x_2,\dots,x_n\},a=x_0<x_1<x_2<\dots <x_n=b$$ of $[a, b]$ and a partition $$Q=\{y_0,y_1,y_2,\dots,y_m\},c=y_0<y_1<y_2<\dots<y_m=d$$ of $[c, d] $. Together these partitions $P, Q$ divide the rectangular region $[a, b] \times [c, d] $ into a finite number (namely $mn$) of smaller rectangular regions of type $[x_{i-1},x_{i}]\times[y_{j-1},y_{j}]$. Out of these small regions there will be some which have a non empty intersection with our given plane region $A$. Choose these smaller rectangles and add their area to form the sum $U(A, P, Q)$ so that $$U(A, P, Q) =\sum_{A\cap([x_{i-1},x_{i}]\times[y_{j-1},y_{j}]) \neq \emptyset} (x_{i} - x_{i-1})(y_{j}-y_{j-1})$$ Clearly this number defined above is positive. Next consider those small rectangular regions which are contained in set $A$ and form the sum $$L(A, P, Q)=\sum_{[x_{i-1},x_{i}]\times [y_{j-1},y_{j}]\subseteq A} (x_{i} - x_{i-1})(y_{j}-y_{j-1})$$ The above sum is clearly non negative. And we can note that $L(A, P, Q) \leq U(A, P, Q) $. These numbers $L, U$ defined above are dependent on the region $A$ as well as partitions $P, Q$. Moreover as we choose different partitions $P, Q$ these numbers may vary but will remain bounded. Let us define numbers $j, J$ by $$J=\inf\, \{U(A, P, Q) \mid P, Q\text{ are partitions of }[a, b], [c, d] \text{ respectively} \} $$ and $$j=\sup\, \{L(A, P, Q) \mid P, Q\text{ are partitions of }[a, b], [c, d] \text{ respectively} \} $$ Both these numbers $j, J$ exist and $j\leq J$. If $j=J$ we say that the plane region $A$ is Jordan measurable and its Jordan content is the common value of $j$ and $J$. We also say that area of plane region $A$ is $j=J$.

When the plane region $A$ is a region under graph of a non-negative function $f$ defined on interval $[a, b]$ so that $$A=\{(x, y) \mid x\in[a, b], y\in[0,f(x)]\} $$ then it can be easily proved that the region $A$ is Jordan measurable if and only if $f$ is Riemann integrable on  $[a, b] $ and its area is $\int_{a} ^{b} f(x) \, dx$.

Interesting things happen when $f$ is continuous on $[a, b]$ and then the Fundamental Theorem of Calculus tells us that there is a function $F$ defined on $[a, b] $ such that $F'(x) =f(x) $ for all $x\in [a, b] $ and moreover the area of the region under graph of $f$ is given by $$\int_{a} ^{b} f(x) \, dx=F(b) - F(a)$$ Thus the answer to your question lies in the rigorous definitions of area and integral and the Fundamental Theorem of Calculus as explained above. I have only tried to provide definitions and avoided proofs to control the length of the answer. Proofs are not too difficult but do require some effort on part of reader. Also understand that the intuitive arguments based on geometrical drawings are not substitute for proofs and only paint a vague picture. 
