# Given Vallias' value for int_{0}^{sqrt{a}} x^{2}dx, calculate his value for int_{0}^{a} sqrt{x}dx, using graph.

ediculeN 2020-11-12 Answered
Given Vallias' value for ${\int }_{0}^{\sqrt{a}}{x}^{2}dx$,
calculate his value for ${\int }_{0}^{a}\sqrt{x}dx,$ using graph.
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## Expert Answer

oppturf
Answered 2020-11-13 Author has 94 answers

Step 1
Let's look at this picture and look at the regions

Step 2
From the above figure, it is observed that length and breadth of the rectangle are a and square root of a. the area of rectangle can be given as follows.
Area of the $={\int }_{0}^{\sqrt{a}}{x}^{2}dx+{\int }_{0}^{a}\sqrt{x}dx$
$length×breadth={\int }_{0}^{\sqrt{a}}{x}^{2}dx+{\int }_{0}^{a}\sqrt{x}dx$
$a×\sqrt{a}={\int }_{0}^{\sqrt{a}}{x}^{2}dx+{\int }_{0}^{a}\sqrt{x}dx$
${\int }_{0}^{a}\sqrt{x}dx=a\sqrt{a}-{\int }_{0}^{\sqrt{a}}{x}^{2}dx$
Thus, for given Wallis's value for ${\int }_{0}^{\sqrt{a}}{x}^{2}dx$, it be obtained that
${\int }_{0}^{a}\sqrt{x}dx=a\sqrt{a}-{\int }_{0}^{\sqrt{a}}{x}^{2}dx$

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