# Thread: find area under the curve using calculus

1. ## find area under the curve using calculus

hi
how are u all
i have a math problem that require from
me to find area under the curve
but because am not good in math i didn't
understand how to find the expresion in the blue
rectangle
in the attached pic

any help
is appreciated ...

secondly ...to make sure i will understand this
type of problems and how to solve it,
could any one plz solve for me the problem
below(find tha area under curve)

2. ## Re: find area under the curve using calculus

There is no 'finding' $(i- 1/n)\Delta x$, that is an arbitrary choice. Once you have divided the interval 0 to 1 in n subintervals, each having length $\Delta x$, and so right endpoints $\Delta x$, $2\Delta x$, $3\Delta x$, ..., $i\Delta x$, ..., you must choose one x in each interval. Here they have just chosen the simplest- the midpoint: $i\Delta x+ (1/2)\Delta x= (i+ 1/2)\Delta x$. You could as easily have chosen $(i+ 1/4)\Delta x$ or $(i+ 1/3)\Delta x$ or $(i+3/4)\Delta x$, etc.

In the last problem, all of the individual areas are rectangles or trapezoids. Do you know the formulas for area of a rectangle or trapezoid? Find the area of each and add them.

3. ## Re: find area under the curve using calculus

Originally Posted by HallsofIvy
There is no 'finding' $(i- 1/n)\Delta x$, that is an arbitrary choice. Once you have divided the interval 0 to 1 in n subintervals, each having length $\Delta x$, and so right endpoints $\Delta x$, $2\Delta x$, $3\Delta x$, ..., $i\Delta x$, ..., you must choose one x in each interval. Here they have just chosen the simplest- the midpoint: $i\Delta x+ (1/2)\Delta x= (i+ 1/2)\Delta x$. You could as easily have chosen $(i+ 1/4)\Delta x$ or $(i+ 1/3)\Delta x$ or $(i+3/4)\Delta x$, etc.

In the last problem, all of the individual areas are rectangles or trapezoids. Do you know the formulas for area of a rectangle or trapezoid? Find the area of each and add them.
hi
However, if you want the exact area under the curve, say $y = x^2$, then you'd need an integral, $\int_{0}^{1} x^2 \,dx$, which is equal to 1/3.