prove that ∫ a b f + ∫ f ( a ) f ( b ) f − 1 = b f ( b ) − a f ( a )Let 0 &lt; a &lt; b let f &gt; 0 be continuous and strictly increasing function on [a,b]. Prove that ∫ a b f + ∫ f ( a ) f ( b ) f − 1 = b f ( b ) − a f ( a ).How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.

Question

prove that       ∫          a        b    f     +      ∫          f      (      a      )              f      (      b      )               f          −      1        =  b  f  (  b  )  −  a  f  (  a  )Let   0  &amp;lt;  a  &amp;lt;  b    let   f  &amp;gt;  0 be continuous and strictly increasing function on [a,b]. Prove that       ∫          a        b    f     +      ∫          f      (      a      )              f      (      b      )               f          −      1        =  b  f  (  b  )  −  a  f  (  a  ).How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.

Carmelo Payne · Accepted Answer

Step 1Add any left Riemann sum with partition   (  a  ,      x    1    ,  …  ,      x          n      −      1        ,  b  ) for the first integral to a right Riemann sum with partition   (  f  (  a  )  ,  f  (      x    1    )  ,  …  ,  f  (      x          n      −      1        )  ,  f  (  b  )  ) for the second integral to get      ∑          j      =      1        n    f  (      x          j      −      1        )  (      x    j    −      x          j      −      1        )  +      ∑          j      =      1        n        f          −      1        (  f  (      x    j    )  )  (  f  (      x    j    )  −  f  (      x          j      −      1        )  )    =      ∑          j      =      1        n    f  (      x          j      −      1        )  (      x    j    −      x          j      −      1        )  +      ∑          j      =      1        n        x    j    (  f  (      x    j    )  −  f  (      x          j      −      1        )  )    =      ∑          j      =      1        n    (        x    j    f  (      x    j    )  −      x          j      −      1        f  (      x          j      −      1        )    )    =  b  f  (  b  )  −  a  f  (  a  )The LHS converges to       ∫    a    b    f  +      ∫          f      (      a      )              f      (      b      )            f          −      1       as the partition norm goes to 0 and the RHS stays the same.

Boilanubjaini8f · Accepted Answer

Think about the strictly growing function&#039;s graph&amp;nbsp;y&amp;nbsp;=&amp;nbsp;f&amp;nbsp;(&amp;nbsp;x&amp;nbsp;). In this graph, the points (a, f(a)) and (b, f(b)) are intersected.The area under the curve&amp;nbsp;y&amp;nbsp;=&amp;nbsp;f&amp;nbsp;(&amp;nbsp;x&amp;nbsp;)&amp;nbsp;from&amp;nbsp;x&amp;nbsp;=&amp;nbsp;a&amp;nbsp;to&amp;nbsp;x&amp;nbsp;=&amp;nbsp;b&amp;nbsp;is given by&amp;nbsp;∫&amp;nbsp;a&amp;nbsp;b&amp;nbsp;f&amp;nbsp;(&amp;nbsp;x&amp;nbsp;)&amp;nbsp;&amp;nbsp;d&amp;nbsp;x&amp;nbsp;. It will be the graph of the inverse function when looking at the same graph &quot;from&quot; the y-axis&amp;nbsp;f&amp;nbsp;−&amp;nbsp;1&amp;nbsp;. The area under the inverse curve from&amp;nbsp;y&amp;nbsp;=&amp;nbsp;f&amp;nbsp;(&amp;nbsp;a&amp;nbsp;)&amp;nbsp;to&amp;nbsp;y&amp;nbsp;=&amp;nbsp;f&amp;nbsp;(&amp;nbsp;b&amp;nbsp;)&amp;nbsp;is given by&amp;nbsp;∫&amp;nbsp;f&amp;nbsp;(&amp;nbsp;a&amp;nbsp;)&amp;nbsp;f&amp;nbsp;(&amp;nbsp;b&amp;nbsp;)&amp;nbsp;f&amp;nbsp;−&amp;nbsp;1&amp;nbsp;(&amp;nbsp;y&amp;nbsp;)&amp;nbsp;&amp;nbsp;d&amp;nbsp;y&amp;nbsp;. The sum of these two areas is&amp;nbsp;∫&amp;nbsp;a&amp;nbsp;b&amp;nbsp;f&amp;nbsp;(&amp;nbsp;x&amp;nbsp;)&amp;nbsp;&amp;nbsp;d&amp;nbsp;x&amp;nbsp;+&amp;nbsp;∫&amp;nbsp;f&amp;nbsp;(&amp;nbsp;a&amp;nbsp;)&amp;nbsp;f&amp;nbsp;(&amp;nbsp;b&amp;nbsp;)&amp;nbsp;f&amp;nbsp;−&amp;nbsp;1&amp;nbsp;(&amp;nbsp;y&amp;nbsp;)&amp;nbsp;&amp;nbsp;d&amp;nbsp;y&amp;nbsp;..Today, the following methods can also be used to locate this area: Take into account the rectangle with two corners at (0,0) and (b,f(b)). This&#039;s area is specified by bf (b). Now if we remove the area af(a) of the smaller rectangle with corners at (0,0) and (a,f(a)) we get&amp;nbsp;b&amp;nbsp;f&amp;nbsp;(&amp;nbsp;b&amp;nbsp;)&amp;nbsp;−&amp;nbsp;a&amp;nbsp;f&amp;nbsp;(&amp;nbsp;a&amp;nbsp;)&amp;nbsp;..The area that we estimated earlier using integrals corresponds to this one. This makes them equal.

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