The definite integral between two points $a$ and $b$ gives the area under the curve for the function between $a$ and $b$, textbooks say.
A question arises - how?
I had this doubt when I started to learn calculus in my school. And I decided to set out to prove it.
Consider a function $\operatorname{f}$, and its derivative $\operatorname{f'}$. We have to prove that
$\int_a^b \operatorname{f'}(x) = $ Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$
Let Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$ = $S$
Divide the shaded region into parts of very very small width $\delta \to 0$. Each part can be considered as a rectangle of width $\delta$ and height $\operatorname{f'}(c)$ where $c$ is the $x$ coordinate of the graph, $a\le c\le b$.
$$\therefore S = \lim_{\delta \to 0}\Bigl(\delta \times \operatorname{f'}(a) + \delta \times \operatorname{f'}(a+\delta) + \delta \times \operatorname{f'}(a+2\delta) +...\\+ \delta \times \operatorname{f'}(b-\delta) + \delta \times \operatorname{f'}(b)\Bigr)$$
From definition,
$$\operatorname{f'}(a+n\delta) = \lim_{\delta \to 0}\frac{\operatorname{f}(a+(n+1)\delta) - \operatorname{f}(a+n\delta)}{\delta}$$
$$\operatorname{f'}(b-n\delta) = \lim_{\delta \to 0}\frac{\operatorname{f}(b-(n-1)\delta) - \operatorname{f}(b-n\delta)}{\delta}$$
Substituting in $S$,
$$S = \lim_{\delta \to 0}\Biggl( \Bigl(\operatorname{f}(a+\delta) - \operatorname{f}(a)\Bigr) + \Bigl(\operatorname{f}(a+2\delta) - \operatorname{f}(a+\delta)\Bigr) + \Bigl(\operatorname{f}(a+3\delta) - \operatorname{f}(a+2\delta)\Bigr) +...\\+ \Bigl(\operatorname{f}(b) - \operatorname{f}(b-\delta)\Bigr) + \Bigl(\operatorname{f}(b+\delta) - \operatorname{f}(b)\Bigr) \Biggr)$$
Cancelling the terms,
$$S = \lim_{\delta \to 0}\Bigl(\operatorname{f}(b+\delta) - \operatorname{f}(a)\Bigr) = \operatorname{f}(b) - \operatorname{f}(a) = \int_a^b \operatorname{f'}(x) $$
Thus we have successfully proved that the definite integral between two points $a$ and $b$ gives the area under the curve for the function between $a$ and $b$.
A question arises - how?
I had this doubt when I started to learn calculus in my school. And I decided to set out to prove it.
Consider a function $\operatorname{f}$, and its derivative $\operatorname{f'}$. We have to prove that
$\int_a^b \operatorname{f'}(x) = $ Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$
Let Area under the curve for $\operatorname{f'}(x)$ between $a$ and $b$ = $S$
Divide the shaded region into parts of very very small width $\delta \to 0$. Each part can be considered as a rectangle of width $\delta$ and height $\operatorname{f'}(c)$ where $c$ is the $x$ coordinate of the graph, $a\le c\le b$.
$$\therefore S = \lim_{\delta \to 0}\Bigl(\delta \times \operatorname{f'}(a) + \delta \times \operatorname{f'}(a+\delta) + \delta \times \operatorname{f'}(a+2\delta) +...\\+ \delta \times \operatorname{f'}(b-\delta) + \delta \times \operatorname{f'}(b)\Bigr)$$
From definition,
$$\operatorname{f'}(a+n\delta) = \lim_{\delta \to 0}\frac{\operatorname{f}(a+(n+1)\delta) - \operatorname{f}(a+n\delta)}{\delta}$$
$$\operatorname{f'}(b-n\delta) = \lim_{\delta \to 0}\frac{\operatorname{f}(b-(n-1)\delta) - \operatorname{f}(b-n\delta)}{\delta}$$
Substituting in $S$,
$$S = \lim_{\delta \to 0}\Biggl( \Bigl(\operatorname{f}(a+\delta) - \operatorname{f}(a)\Bigr) + \Bigl(\operatorname{f}(a+2\delta) - \operatorname{f}(a+\delta)\Bigr) + \Bigl(\operatorname{f}(a+3\delta) - \operatorname{f}(a+2\delta)\Bigr) +...\\+ \Bigl(\operatorname{f}(b) - \operatorname{f}(b-\delta)\Bigr) + \Bigl(\operatorname{f}(b+\delta) - \operatorname{f}(b)\Bigr) \Biggr)$$
Cancelling the terms,
$$S = \lim_{\delta \to 0}\Bigl(\operatorname{f}(b+\delta) - \operatorname{f}(a)\Bigr) = \operatorname{f}(b) - \operatorname{f}(a) = \int_a^b \operatorname{f'}(x) $$
Thus we have successfully proved that the definite integral between two points $a$ and $b$ gives the area under the curve for the function between $a$ and $b$.
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