calculate the following area. first draw a neat sketch indicating the required area. y= 5x + 6 - x^2 for which y is positive

Question

The area enclosed by the x-axis and that part of the curve

Answer 1

y=5x+6-x^2=(1+x)(6-x) so curve cuts x axis at -1 and 6. The limits for the area are therefore -1 to 6.

The area is given by S[-1,6]((5x+6-x^2)dx) where S denotes integral. The area is the sum of the areas of thin rectangles with height y and width dx. The area of each rectangle is ydx.

The integral is (5x^2/2+6x-x^3/3)[-1,6]=(90+36-72)-(5/2+6-1/3)=54-49/6=45.833 approx. This is the area under the curve between the curve and the x axis.

 

Answer 2

Given the formula of the parabola,
Say it is 5x+6-x^2 it´s primitive integral is 6x+5x^2/2-x^3/3 so if we need the area below the parabola, say between x=-1 and x=6

The area is 343/6 = 57.1667

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