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Integration Between Axes

0

2

$begingroup$

The diagram shows the graph of $y=x^2$, where $ain[1,∞]$. The area of the pink region is equal to the area of the blue region. Give two equations for $a$ in terms of $b$, and hence give $a$ in exact form and determine the size of the blue area.

So my first instinct was to take the integral of both to find the areas and say that $ab-1$ the area of the entire rectangle there minus the $1$ by $1$, is equal to $(frac{a^3}{3}-frac{1}{3})+(frac{2}{3}b^frac{3}{2}-frac{2}{3})$
Of course this didn't really get me anywhere because it turned out to just be an identity. Any ideas?

integration geometry definite-integrals

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asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

$endgroup$

add a comment | 

0

2

$begingroup$

The diagram shows the graph of $y=x^2$, where $ain[1,∞]$. The area of the pink region is equal to the area of the blue region. Give two equations for $a$ in terms of $b$, and hence give $a$ in exact form and determine the size of the blue area.

So my first instinct was to take the integral of both to find the areas and say that $ab-1$ the area of the entire rectangle there minus the $1$ by $1$, is equal to $(frac{a^3}{3}-frac{1}{3})+(frac{2}{3}b^frac{3}{2}-frac{2}{3})$
Of course this didn't really get me anywhere because it turned out to just be an identity. Any ideas?

integration geometry definite-integrals

share|cite|improve this question

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

$endgroup$

add a comment | 

$begingroup$

The diagram shows the graph of $y=x^2$, where $ain[1,∞]$. The area of the pink region is equal to the area of the blue region. Give two equations for $a$ in terms of $b$, and hence give $a$ in exact form and determine the size of the blue area.

So my first instinct was to take the integral of both to find the areas and say that $ab-1$ the area of the entire rectangle there minus the $1$ by $1$, is equal to $(frac{a^3}{3}-frac{1}{3})+(frac{2}{3}b^frac{3}{2}-frac{2}{3})$
Of course this didn't really get me anywhere because it turned out to just be an identity. Any ideas?

integration geometry definite-integrals

share|cite|improve this question

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

$endgroup$

share|cite|improve this question

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

share|cite|improve this question

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

asked Mar 26 at 0:23

Savvas NicolaouSavvas Nicolaou

867

3 Answers
3

active

oldest

votes

2

$begingroup$

Hints

(1) Where have you used the fact that the blue and pink areas are equal?

(2) You also know that $b=a^2$ since $(a,b)$ lies on the graph.

share|cite|improve this answer

answered Mar 26 at 0:29

MPWMPW

31.2k12157

$endgroup$

add a comment | 

1

$begingroup$

You should notice the following two facts:

The area of the regions are the same

$intlimits_1^a x^2~dx = intlimits_1^b sqrt{y}~dy$

The vertical line for $x=a$ intersects the curve at the same point as the horizontal line for $y=b$

$b = a^2$

Now, perform some calculus and some algebra to find the exact values of each.

share|cite|improve this answer

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I get for a in my second equation $a=sqrt[3]{2b^frac{3}{2}+1}$
  $endgroup$
  – Savvas Nicolaou
  Mar 26 at 0:33
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Can someone try to solve it? I keep getting an answer a = 1, which is incorrect according to the mark scheme (and well, it also doesn't fit the domain).

share|cite|improve this answer

answered Apr 3 at 18:35

MarcinMarcin

83

$endgroup$

add a comment | 

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2

$begingroup$

Hints

(1) Where have you used the fact that the blue and pink areas are equal?

(2) You also know that $b=a^2$ since $(a,b)$ lies on the graph.

share|cite|improve this answer

answered Mar 26 at 0:29

MPWMPW

31.2k12157

$endgroup$

add a comment | 

2

$begingroup$

Hints

(1) Where have you used the fact that the blue and pink areas are equal?

(2) You also know that $b=a^2$ since $(a,b)$ lies on the graph.

share|cite|improve this answer

answered Mar 26 at 0:29

MPWMPW

31.2k12157

$endgroup$

add a comment | 

$begingroup$

Hints

(1) Where have you used the fact that the blue and pink areas are equal?

(2) You also know that $b=a^2$ since $(a,b)$ lies on the graph.

share|cite|improve this answer

answered Mar 26 at 0:29

MPWMPW

31.2k12157

$endgroup$

share|cite|improve this answer

answered Mar 26 at 0:29

MPWMPW

31.2k12157

answered Mar 26 at 0:29

MPWMPW

31.2k12157

answered Mar 26 at 0:29

MPWMPW

31.2k12157

1

$begingroup$

You should notice the following two facts:

The area of the regions are the same

$intlimits_1^a x^2~dx = intlimits_1^b sqrt{y}~dy$

The vertical line for $x=a$ intersects the curve at the same point as the horizontal line for $y=b$

$b = a^2$

Now, perform some calculus and some algebra to find the exact values of each.

share|cite|improve this answer

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I get for a in my second equation $a=sqrt[3]{2b^frac{3}{2}+1}$
  $endgroup$
  – Savvas Nicolaou
  Mar 26 at 0:33
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You should notice the following two facts:

The area of the regions are the same

$intlimits_1^a x^2~dx = intlimits_1^b sqrt{y}~dy$

The vertical line for $x=a$ intersects the curve at the same point as the horizontal line for $y=b$

$b = a^2$

Now, perform some calculus and some algebra to find the exact values of each.

share|cite|improve this answer

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So would I get for a in my second equation $a=sqrt[3]{2b^frac{3}{2}+1}$
  $endgroup$
  – Savvas Nicolaou
  Mar 26 at 0:33
  
  
  

  
  
  
  

add a comment | 

$begingroup$

You should notice the following two facts:

The area of the regions are the same

$intlimits_1^a x^2~dx = intlimits_1^b sqrt{y}~dy$

The vertical line for $x=a$ intersects the curve at the same point as the horizontal line for $y=b$

$b = a^2$

Now, perform some calculus and some algebra to find the exact values of each.

share|cite|improve this answer

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

$endgroup$

share|cite|improve this answer

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

answered Mar 26 at 0:29

JMoravitzJMoravitz

49.4k44091

0

$begingroup$

Can someone try to solve it? I keep getting an answer a = 1, which is incorrect according to the mark scheme (and well, it also doesn't fit the domain).

share|cite|improve this answer

answered Apr 3 at 18:35

MarcinMarcin

83

$endgroup$

add a comment | 

0

$begingroup$

Can someone try to solve it? I keep getting an answer a = 1, which is incorrect according to the mark scheme (and well, it also doesn't fit the domain).

share|cite|improve this answer

answered Apr 3 at 18:35

MarcinMarcin

83

$endgroup$

add a comment | 

$begingroup$

Can someone try to solve it? I keep getting an answer a = 1, which is incorrect according to the mark scheme (and well, it also doesn't fit the domain).

share|cite|improve this answer

answered Apr 3 at 18:35

MarcinMarcin

83

$endgroup$

share|cite|improve this answer

answered Apr 3 at 18:35

MarcinMarcin

83

answered Apr 3 at 18:35

MarcinMarcin

83

answered Apr 3 at 18:35

MarcinMarcin

83