Explanation Given: The area between the x -axis and nonnegative function s ( x ) d r d x = h ( x ) is 5 on the interval [ 0 , 2 ] with r ( 0 ) = 0 . On the same interval the area between the x -axis and nonnegative function r ( x ) d s d x = k ( x ) is 10. Formula used: “If u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u .” Calculation: Rearrange the formula and obtain, u v = ∫ v d u + ∫ u d v . Then, the above formula can be rearranged in terms of r ( x ) and s ( x ) as, ∫ r ( x ) d s + ∫ s ( x ) d r = r ( x ) s ( x ) The area between the x- axis and the nonnegative function h ( x ) over the interval [ 0 , 2 ] is 5. That is ∫ 0 2 h ( x ) d x = 5 and ∫ 0 2 h ( x ) d r = 5 . The area between the x axis and the nonnegative function k ( x ) over the interval [ 0 , 2 ] is 10

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Chapter 8.1, Problem 33E

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The value of the function r(x)s(x) at x=2 .

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Chapter 8.1, Problem 32E

Chapter 8.1, Problem 34E

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The value of the function r ( x ) s ( x ) at x = 2 .

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The value of the function r(x)s(x) at x=2 .

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