# How do you find the integral #int_1^(3)12(1+5x)^5dx# ?

However, if one isn't yet familiar with the change of variable rule, one can just as easily integrate this particular function rather easily (albeit slightly more prone to error) by expanding the brackets and integrating a particularly long function.

The plus side of the second method is that it's easy, and it can be used to check that you have your change-of-variable rule method correct. The minus side is that it's more prone to errors, and takes up more time (which is valuable in tests and exams).

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To find the integral of ( \int_{1}^{3} 12(1+5x)^5 , dx ), you can use the power rule for integration and then evaluate the integral at the upper and lower limits.

First, use the power rule to integrate ( (1+5x)^5 ) with respect to ( x ), which results in ( \frac{1}{6}(1+5x)^6 ).

Next, multiply the integrated function by 12, yielding ( 12 \times \frac{1}{6}(1+5x)^6 = 2(1+5x)^6 ).

Now, evaluate this expression at the upper limit (3) and subtract the value at the lower limit (1):

[ \left[ 2(1+5x)^6 \right]_{1}^{3} ]

[ = 2(1+5(3))^6 - 2(1+5(1))^6 ]

[ = 2(16^6) - 2(6^6) ]

[ = 2 \times 16777216 - 2 \times 46656 ]

[ = 33554432 - 93312 ]

[ = 33461120 ]

So, ( \int_{1}^{3} 12(1+5x)^5 , dx = 33461120 ).

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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.

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.

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.

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