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What is #f(x) = int xsqrt(x^2-2) dx# if #f(3) = 3 #?

Answer 1

#f(x)=1/3 (x^2-2)^(3/2) +3-(7sqrt7)/3#

#f(x)=intxsqrt(x^2-2)dx=intx(x^2-2)^(1/2)dx#

Note that the x outside of the square root is part of the derivative of the inside so we can just integrate #(x^2-2)^(1/2)#

#f(x)=intx(x^2-2)^(1/2)dx = 2/3 (x^2-2)^(3/2)*1/2#-> since the 2 from the derivative of the inside is missing.

#f(x)=1/3 (x^2-2)^(3/2) +C# #f(3)=1/3 (9-2)^(3/2)+C=3# #(7sqrt7)/3 +C =3# #C=3-(7sqrt7)/3# #f(x)=1/3 (x^2-2)^(3/2) +3-(7sqrt7)/3#

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Answer 2

Christopher Allers

The integral of xsqrt(x^2 - 2) dx equals (1/3)(x^2 - 2)^(3/2) + C. Given f(3) = 3, we can solve for C, resulting in C = 0. Thus, the integral evaluates to (1/3)(x^2 - 2)^(3/2).

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Answer 3

Emily Ammerman

To find the function ( f(x) = \int x \sqrt{x^2 - 2} , dx ) when ( f(3) = 3 ), you need to evaluate the definite integral using the given value. First, find the indefinite integral of the function. Then, use the given information to determine the constant of integration. Finally, evaluate the function at ( x = 3 ) to find its value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.