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Find F'(x) when F(x)= the integral of (1/t^6)dt on the interval [2,x^2].

We have the problem ∫1/t^6 dt on the interval [2,x²].

The problem is really ∫t^(-6) dt on the interval [2,x²].

The answer is found by adding one to the exponent and dividing by the new exponent.

That gives t^(-5)/(-5) evaluated from 2 to x².

That becomes [x²^(-5) - 2^(-5)]/(-5).

When taking a square to another power, multiply the exponents.

When taking 2^5, 32 is gotten. When taking to the -5, 1/32 is then gotten.

Switch the two terms in the numerator to eliminate the negative sign in the denominator.

That gives [1/32 - 1/x^10]/5.

This can be rewritten by combining the two fractions.

This gives (x^10 - 32)/(160x^10).

Calculus

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