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(d). Find the area of the region enclosed by the graphs of f(x)= x³ and g(x)=x²+2x. [Verify your answer by MATHEMATICA and attach the printout of the commands and output (e). Find the area of the region enclosed by the graphs of y = 3/X and y=4-X. [Verify your answer by MATHEMATICA and attach the printout of the commands and output

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Answer:

d .  37/12 unit^2.

Step-by-step explanation:

d) First find the points of intersection of the the 2 graphs, by solving them simultaneously:

x^3 = x^2 + 2x

x^3 - x^2 - 2x) = 0

x(x^2 - x - 2) = 0

x(x - 2)(x + 1) = 0

So they intersect at x = -1, x = 0 and x = 2.

Take the area from x = 0 to x = 2)

           2                        2

Area =  ∫ x^2 + 2x dx -   ∫  x^3 dx

            0                       0

   2

=  [ x^3/3 + x^2 ) - x^4/4]

   0

=   ((8/3 + 4) - 0) - (16/4 - 0) -

=   8/3 unit^2

Now calculate the area  from x = -1 to x = 0.

   0

=    (x^3/3 + x^2 ) - x^4/4 )

   -1

=   (( 0 +1/3 - 1) ) - (0 - 1/4)

=  5/12 unit^2

So the total area of the region =  8/3 + 5/12 = 37/12 unit^2.

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