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Evaluate the integral by interpreting it in terms of area..? where did the (1/4)π(7)^2 come from ?

integral from -7 to 0 (5+ sqrt(49-x^2)) dx

in this question here where did the (1/4)π(7)^2 come from ? im talking about the one thats the best answer..thanks

http://answers.yahoo.com/question/index?qid=201111...

Note that you can split up the integral to get:

∫ [5 + √(49 - x^2)] dx (from x=-7 to 0)

= ∫ 5 dx (from x=-7 to 0) + ∫ √(49 - x^2) dx (from x=-7 to 0).

The first integral represents the area under y = 5 on the interval (-7, 0), which is a rectangle of width 7 and height 5. This rectangle has area (5)(7) = 35, so:

∫ 5 dx (from x=-7 to 0) = 35.

The second integrand is y = √(49 - x^2), which is the upper semi-circle centered at the origin with a radius of 7. The interval (-7, 0) represents the left half of the semi-circle, so ∫ √(49 - x^2) dx (from x=-7 to 0) just represents the area of a quarter circle of radius 7. Thus:

∫ √(49 - x^2) dx = (1/4)π(7)^2 = 49π/4.

Therefore, adding these two results together gives the required value of the integral to be 35 + 49π/4.

I hope this helps!

1 year ago

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