int(2 to 0) x^2 dx
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I think you mean 0 to 2
Integral
= ∫ (x = 0 to 2) x^2 dx
= (1/3) [x^3] ... (x = 0 to 2)
= (1/3) (2)^3 - 0
= 8/3.
Integral
= ∫ (x = 0 to 2) x^2 dx
= (1/3) [x^3] ... (x = 0 to 2)
= (1/3) (2)^3 - 0
= 8/3.
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The triviality and form of this expression is suspect. It gives me cause to wonder whether youre interested in solving the integral via an infinite Riemann sum or by more advanced means. Im compelled to ask, What is your current level of understanding?
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Int 2 --> 0 x^2 dx
soln:
(x^3)/2 ] limit 2 --> 0
then substituting the lower and upper limits we get
=1/3 [0 - 2^3]
=1/3 [-8]
= - 8/3
soln:
(x^3)/2 ] limit 2 --> 0
then substituting the lower and upper limits we get
=1/3 [0 - 2^3]
=1/3 [-8]
= - 8/3