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1

$begingroup$

I tried to ask this in a different way and did not correctly explain myself.

I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.

Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.

I did not do a good job explaining this on my previous question. Sorry

calculus integration

share|cite|improve this question

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

$endgroup$

add a comment | 

1

$begingroup$

I tried to ask this in a different way and did not correctly explain myself.

I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.

Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.

I did not do a good job explaining this on my previous question. Sorry

calculus integration

share|cite|improve this question

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

$endgroup$

add a comment | 

$begingroup$

I tried to ask this in a different way and did not correctly explain myself.

I am ok integrating the line $y = x$ , let us say from $0$ to $2$ using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.

Here is my question. If I use a straight line above the x axis my equation becomes $y$ = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have one $x$ term and it is not a square. The answers match OK it's the dimensions that bother me.

I did not do a good job explaining this on my previous question. Sorry

calculus integration

share|cite|improve this question

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

$endgroup$

share|cite|improve this question

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

share|cite|improve this question

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

edited Mar 19 at 19:48

ADITYA PRAKASH

365110

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

asked Mar 19 at 16:00

SedumjoySedumjoy

658316

2 Answers
2

active

oldest

votes

3

$begingroup$

The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.

Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.

$$begin{aligned}text{Area }&=lim_{nto infty}sum_{i=1}^{n}underbrace{k}_{text{height}}cdotunderbrace{left(dfrac{b-a}{n}right)}_{text{width of each infinitesimal rectangle}}\&=int_{a}^{b}underbrace{k}_{text{units}}underbrace{mathrm dx}_{text{units}}\&=kxbiggr|_{a}^{b}=k(b-a) text{ sq. units}end{aligned}$$

share|cite|improve this answer

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

$endgroup$

add a comment | 

1

$begingroup$

Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.

Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.

The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.

The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,

$$int_{x=a}^{x=b}ydx=int_{x=a}^{x=b}f(x)dx=int_{x=a}^{x=b}Kdx=K(b-a)$$
Hope this helps...

share|cite|improve this answer

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

$endgroup$

add a comment | 

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3

$begingroup$

The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.

Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.

$$begin{aligned}text{Area }&=lim_{nto infty}sum_{i=1}^{n}underbrace{k}_{text{height}}cdotunderbrace{left(dfrac{b-a}{n}right)}_{text{width of each infinitesimal rectangle}}\&=int_{a}^{b}underbrace{k}_{text{units}}underbrace{mathrm dx}_{text{units}}\&=kxbiggr|_{a}^{b}=k(b-a) text{ sq. units}end{aligned}$$

share|cite|improve this answer

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

$endgroup$

add a comment | 

3

$begingroup$

The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.

Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.

$$begin{aligned}text{Area }&=lim_{nto infty}sum_{i=1}^{n}underbrace{k}_{text{height}}cdotunderbrace{left(dfrac{b-a}{n}right)}_{text{width of each infinitesimal rectangle}}\&=int_{a}^{b}underbrace{k}_{text{units}}underbrace{mathrm dx}_{text{units}}\&=kxbiggr|_{a}^{b}=k(b-a) text{ sq. units}end{aligned}$$

share|cite|improve this answer

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

$endgroup$

add a comment | 

$begingroup$

The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is $k$ and width is $b-a$.

Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.

$$begin{aligned}text{Area }&=lim_{nto infty}sum_{i=1}^{n}underbrace{k}_{text{height}}cdotunderbrace{left(dfrac{b-a}{n}right)}_{text{width of each infinitesimal rectangle}}\&=int_{a}^{b}underbrace{k}_{text{units}}underbrace{mathrm dx}_{text{units}}\&=kxbiggr|_{a}^{b}=k(b-a) text{ sq. units}end{aligned}$$

share|cite|improve this answer

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

answered Mar 19 at 16:18

Paras KhoslaParas Khosla

2,867523

1

$begingroup$

Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.

Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.

The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.

The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,

$$int_{x=a}^{x=b}ydx=int_{x=a}^{x=b}f(x)dx=int_{x=a}^{x=b}Kdx=K(b-a)$$
Hope this helps...

share|cite|improve this answer

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

$endgroup$

add a comment | 

1

$begingroup$

Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.

Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.

The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.

The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,

$$int_{x=a}^{x=b}ydx=int_{x=a}^{x=b}f(x)dx=int_{x=a}^{x=b}Kdx=K(b-a)$$
Hope this helps...

share|cite|improve this answer

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

$endgroup$

add a comment | 

$begingroup$

Consider a rectangle formed by the equation $y=K$ extending from $x=a$ to $x=b$.

Calculate the area of a small rectangular strip formed by the coordinates $(x,0)$, $(x+dx,0)$,$(x+dx,y)$ and $(x,y)$.

The area of this rectangle will be, $ydx$. Convert the whole rectangular area into smaller rectangles of area $ydx$ and sum all of them from $x=a$ to $x=b$ using integration.

The area under the curve $y=f(x)$ bounded by the X Axis and the lines $x=a$ and $x=b$ is given by,

$$int_{x=a}^{x=b}ydx=int_{x=a}^{x=b}f(x)dx=int_{x=a}^{x=b}Kdx=K(b-a)$$
Hope this helps...

share|cite|improve this answer

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

$endgroup$

share|cite|improve this answer

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110

answered Mar 19 at 17:34

SNEHIL SANYALSNEHIL SANYAL

658110