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How do you find the definite integral for: #((x^(6))dx# for the intervals #[b, 2b]#?

Answer 1

Luke Alvarez

#127/7 b^7#

#int_b^(2b)x^6dx=[1/7x^7]_b^(2b)#

Note that this antiderivative was obtained using the rule #int_a^bx^ndx=[1/(n+1)x^(n+1)]_a^b#.

#=1/7[(2b)^7-b^7]=1/7(128b^7-b^7)=1/7(127b^7)#

#rArrint_b^(2b)x^6dx=127/7 b^7#

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Answer 2

Aurora Alvarado

To find the definite integral of ( x^6 ) over the interval ([b, 2b]), you first need to find the antiderivative of ( x^6 ), which is ( \frac{1}{7}x^7 ). Then, you evaluate this antiderivative at the upper and lower bounds of the interval and subtract the lower bound value from the upper bound value. So the definite integral is:

[ \left[ \frac{1}{7}x^7 \right]_{b}^{2b} = \frac{1}{7}(2b)^7 - \frac{1}{7}b^7 ]

Simplifying this expression gives:

[ \frac{128}{7}b^7 - \frac{1}{7}b^7 = \frac{127}{7}b^7 ]

Therefore, the definite integral of ( x^6 ) over the interval ([b, 2b]) is ( \frac{127}{7}b^7 ).

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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.