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Finding the area under the curve from a graph

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3

0

$begingroup$

What is the area of the shared region in the figure above that is bounded by the $x$-axis and the curve with the equation $y=xsqrt1-x^2$?

This is the problem I was given. I assumed the answer was $0$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $frac13$ and they are only asking for the area above the $x$-axis. That was also incorrect. The answer that was given as correct was: $frac23$. I assume that is because it is the area of the top - which is $frac13$ and the area of the bottom also $frac13$ - which makes $frac23$. But, why is the answer not zero? Doesn't integration count area under the $x$-axis as negative? Does the wording of the question say otherwise?

calculus integration definite-integrals area substitution

share|cite|improve this question

edited Aug 1 at 21:49

Michael Rozenberg

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asked Aug 1 at 20:38

burtburt

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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think an area can be negative? That's like saying that the distance between the points $4$ and $3$ is $-1$ beacuse "you go back". The area of a curve is the integral of the absolute value of the function.
  $endgroup$
  – Alfredo
  Aug 1 at 20:42
  
  

  
  
  
  

add a comment | 

3

0

$begingroup$

What is the area of the shared region in the figure above that is bounded by the $x$-axis and the curve with the equation $y=xsqrt1-x^2$?

This is the problem I was given. I assumed the answer was $0$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $frac13$ and they are only asking for the area above the $x$-axis. That was also incorrect. The answer that was given as correct was: $frac23$. I assume that is because it is the area of the top - which is $frac13$ and the area of the bottom also $frac13$ - which makes $frac23$. But, why is the answer not zero? Doesn't integration count area under the $x$-axis as negative? Does the wording of the question say otherwise?

calculus integration definite-integrals area substitution

share|cite|improve this question

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think an area can be negative? That's like saying that the distance between the points $4$ and $3$ is $-1$ beacuse "you go back". The area of a curve is the integral of the absolute value of the function.
  $endgroup$
  – Alfredo
  Aug 1 at 20:42
  
  

  
  
  
  

add a comment | 

$begingroup$

What is the area of the shared region in the figure above that is bounded by the $x$-axis and the curve with the equation $y=xsqrt1-x^2$?

This is the problem I was given. I assumed the answer was $0$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $frac13$ and they are only asking for the area above the $x$-axis. That was also incorrect. The answer that was given as correct was: $frac23$. I assume that is because it is the area of the top - which is $frac13$ and the area of the bottom also $frac13$ - which makes $frac23$. But, why is the answer not zero? Doesn't integration count area under the $x$-axis as negative? Does the wording of the question say otherwise?

calculus integration definite-integrals area substitution

share|cite|improve this question

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

$endgroup$

share|cite|improve this question

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

share|cite|improve this question

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

edited Aug 1 at 21:49

Michael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

asked Aug 1 at 20:38

burtburt

3771010 bronze badges

4 Answers
4

active

oldest

votes

5

$begingroup$

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).

Note that Area is absolute, but Net Area is relative

share|cite|improve this answer

answered Aug 1 at 20:56

NazimJNazimJ

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$endgroup$

add a comment | 

2

$begingroup$

The integral $int_-1^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $int_0^1y~dx-int_-1^0y~dx$.

share|cite|improve this answer

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

The function is odd, the area you look for is
$$A=2int_0^1xsqrt1-x^2dx$$

put $y=x^2$.

then

$$A=int_0^1sqrt1-ydy$$
$$=int_0^1(1-y)^frac 12dy$$

$$=Bigl[ frac 23(1-y)^frac 32Bigr]_1^0$$
$$=frac 23.$$

share|cite|improve this answer

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Let $x=sin t$.

Thus, we need to get $$2intlimits_0^1xsqrt1-x^2dx=2intlimits_0^fracpi2sintcos^2tdt=intlimits_0^fracpi2sin2tcost=$$
$$=frac12intlimits_0^fracpi2(sin3t+sint)dt=-frac16cos3t-frac12costbig_0^fracpi2=frac23.$$

share|cite|improve this answer

edited Aug 1 at 21:07

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

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$endgroup$

add a comment | 

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5

$begingroup$

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).

Note that Area is absolute, but Net Area is relative

share|cite|improve this answer

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).

Note that Area is absolute, but Net Area is relative

share|cite|improve this answer

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

$endgroup$

add a comment | 

$begingroup$

In your question, it asks for the area of the shaded region, and area is always positive.

For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).

Note that Area is absolute, but Net Area is relative

share|cite|improve this answer

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

answered Aug 1 at 20:56

NazimJNazimJ

1,50422 silver badges1010 bronze badges

2

$begingroup$

The integral $int_-1^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $int_0^1y~dx-int_-1^0y~dx$.

share|cite|improve this answer

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

The integral $int_-1^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $int_0^1y~dx-int_-1^0y~dx$.

share|cite|improve this answer

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

The integral $int_-1^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $int_0^1y~dx-int_-1^0y~dx$.

share|cite|improve this answer

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

answered Aug 1 at 20:43

Shubham JohriShubham Johri

6,93999 silver badges1818 bronze badges

2

$begingroup$

The function is odd, the area you look for is
$$A=2int_0^1xsqrt1-x^2dx$$

put $y=x^2$.

then

$$A=int_0^1sqrt1-ydy$$
$$=int_0^1(1-y)^frac 12dy$$

$$=Bigl[ frac 23(1-y)^frac 32Bigr]_1^0$$
$$=frac 23.$$

share|cite|improve this answer

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

The function is odd, the area you look for is
$$A=2int_0^1xsqrt1-x^2dx$$

put $y=x^2$.

then

$$A=int_0^1sqrt1-ydy$$
$$=int_0^1(1-y)^frac 12dy$$

$$=Bigl[ frac 23(1-y)^frac 32Bigr]_1^0$$
$$=frac 23.$$

share|cite|improve this answer

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

$endgroup$

add a comment | 

$begingroup$

The function is odd, the area you look for is
$$A=2int_0^1xsqrt1-x^2dx$$

put $y=x^2$.

then

$$A=int_0^1sqrt1-ydy$$
$$=int_0^1(1-y)^frac 12dy$$

$$=Bigl[ frac 23(1-y)^frac 32Bigr]_1^0$$
$$=frac 23.$$

share|cite|improve this answer

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

answered Aug 1 at 20:48

hamam_Abdallahhamam_Abdallah

39.1k22 gold badges1616 silver badges3434 bronze badges

2

$begingroup$

Let $x=sin t$.

Thus, we need to get $$2intlimits_0^1xsqrt1-x^2dx=2intlimits_0^fracpi2sintcos^2tdt=intlimits_0^fracpi2sin2tcost=$$
$$=frac12intlimits_0^fracpi2(sin3t+sint)dt=-frac16cos3t-frac12costbig_0^fracpi2=frac23.$$

share|cite|improve this answer

edited Aug 1 at 21:07

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Let $x=sin t$.

Thus, we need to get $$2intlimits_0^1xsqrt1-x^2dx=2intlimits_0^fracpi2sintcos^2tdt=intlimits_0^fracpi2sin2tcost=$$
$$=frac12intlimits_0^fracpi2(sin3t+sint)dt=-frac16cos3t-frac12costbig_0^fracpi2=frac23.$$

share|cite|improve this answer

edited Aug 1 at 21:07

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $x=sin t$.

Thus, we need to get $$2intlimits_0^1xsqrt1-x^2dx=2intlimits_0^fracpi2sintcos^2tdt=intlimits_0^fracpi2sin2tcost=$$
$$=frac12intlimits_0^fracpi2(sin3t+sint)dt=-frac16cos3t-frac12costbig_0^fracpi2=frac23.$$

share|cite|improve this answer

edited Aug 1 at 21:07

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 1 at 21:07

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges

answered Aug 1 at 21:03

Michael RozenbergMichael Rozenberg

124k2020 gold badges105105 silver badges211211 bronze badges