Integral of a modulus function vs absolute value of definite integral












3












$begingroup$


Please refer to the diagram below.
enter image description here



The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.



Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.



I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
enter image description here



I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.



enter image description here



Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...






calculus integration definite-integrals







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$endgroup$












  • $begingroup$
    Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
    $endgroup$
    – Michael Burr
    Jan 2 '17 at 14:52
















3












$begingroup$


Please refer to the diagram below.
enter image description here



The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.



Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.



I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
enter image description here



I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.



enter image description here



Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...






calculus integration definite-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
    $endgroup$
    – Michael Burr
    Jan 2 '17 at 14:52














3












3








3





$begingroup$


Please refer to the diagram below.
enter image description here



The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.



Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.



I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
enter image description here



I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.



enter image description here



Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...






calculus integration definite-integrals







share|cite|improve this question









$endgroup$




Please refer to the diagram below.
enter image description here



The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.



Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.



I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
enter image description here



I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.



enter image description here



Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...






calculus integration definite-integrals




calculus integration definite-integrals






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asked Jan 2 '17 at 14:44









Charlz97Charlz97

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12312












  • $begingroup$
    Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
    $endgroup$
    – Michael Burr
    Jan 2 '17 at 14:52


















  • $begingroup$
    Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
    $endgroup$
    – Michael Burr
    Jan 2 '17 at 14:52
















$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52




$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52










1 Answer
1






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$begingroup$

Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$






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    1 Answer
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    0












    $begingroup$

    Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
    $$
    int_{pi/3}^{2pi/3}|f(x)|,dx
    =
    int_{pi/3}^{2pi/3}|f(x)|,dx
    =
    int_{pi/3}^{pi/2}f(x),dx
    -
    int_{pi/2}^{2pi/3}f(x),dx
    $$
    On the other hand
    $$
    left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
    $$
    does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
    $$
    int_{pi/3}^{2pi/3}|f(x)|,dx=
    (F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
    -F(2pi/3)-F(pi/2)+2F(pi/2)
    $$
    whereas
    $$
    left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
    |F(2pi/3)-F(pi/3)|
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
      $$
      int_{pi/3}^{2pi/3}|f(x)|,dx
      =
      int_{pi/3}^{2pi/3}|f(x)|,dx
      =
      int_{pi/3}^{pi/2}f(x),dx
      -
      int_{pi/2}^{2pi/3}f(x),dx
      $$
      On the other hand
      $$
      left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
      $$
      does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
      $$
      int_{pi/3}^{2pi/3}|f(x)|,dx=
      (F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
      -F(2pi/3)-F(pi/2)+2F(pi/2)
      $$
      whereas
      $$
      left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
      |F(2pi/3)-F(pi/3)|
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
        $$
        int_{pi/3}^{2pi/3}|f(x)|,dx
        =
        int_{pi/3}^{2pi/3}|f(x)|,dx
        =
        int_{pi/3}^{pi/2}f(x),dx
        -
        int_{pi/2}^{2pi/3}f(x),dx
        $$
        On the other hand
        $$
        left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
        $$
        does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
        $$
        int_{pi/3}^{2pi/3}|f(x)|,dx=
        (F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
        -F(2pi/3)-F(pi/2)+2F(pi/2)
        $$
        whereas
        $$
        left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
        |F(2pi/3)-F(pi/3)|
        $$






        share|cite|improve this answer









        $endgroup$



        Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
        $$
        int_{pi/3}^{2pi/3}|f(x)|,dx
        =
        int_{pi/3}^{2pi/3}|f(x)|,dx
        =
        int_{pi/3}^{pi/2}f(x),dx
        -
        int_{pi/2}^{2pi/3}f(x),dx
        $$
        On the other hand
        $$
        left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
        $$
        does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
        $$
        int_{pi/3}^{2pi/3}|f(x)|,dx=
        (F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
        -F(2pi/3)-F(pi/2)+2F(pi/2)
        $$
        whereas
        $$
        left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
        |F(2pi/3)-F(pi/3)|
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 '17 at 14:58









        egregegreg

        180k1485202




        180k1485202






























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