How do you evaluate the definite integral #int (3abs(x3))dx# from [1,4]?
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To evaluate the definite integral ( \int_{1}^{4} (3  x  3) , dx ), we split the integral into two parts based on the absolute value function.

For ( 1 \leq x \leq 3 ): [ 3  x  3 = 3  (3  x) = x ]

For ( 3 < x \leq 4 ): [ 3  x  3 = 3  (x  3) = 6  x ]
Then, we integrate each part separately within the given limits:

For ( 1 \leq x \leq 3 ): [ \int_{1}^{3} x , dx = \left[ \frac{x^2}{2} \right]_{1}^{3} = \frac{9}{2}  \frac{1}{2} = 4 ]

For ( 3 < x \leq 4 ): [ \int_{3}^{4} (6  x) , dx = \left[ 6x  \frac{x^2}{2} \right]_{3}^{4} = (24  8)  (18  4.5) = 16  13.5 = 2.5 ]
Finally, summing up the results from both intervals: [ \int_{1}^{4} (3  x  3) , dx = 4 + 2.5 = 6.5 ]
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To evaluate the definite integral ∫(3  x  3) dx from 1 to 4, you need to break the interval [1, 4] into two parts: from 1 to 3 and from 3 to 4. In each interval, the absolute value term changes sign at x = 3, so you'll have different expressions for the function in each part.
For the interval [1, 3], the function 3  x  3 simplifies to 3  (3  x) = x.
For the interval [3, 4], the function 3  x  3 simplifies to 3  (x  3) = 6  x.
Now, integrate each part separately: ∫(x) dx from 1 to 3 = [x^2 / 2] evaluated from 1 to 3 = (9 / 2)  (1 / 2) = 4
∫(6  x) dx from 3 to 4 = [6x  (x^2 / 2)] evaluated from 3 to 4 = (24  8)  ((18  9) / 2) = 16  (9 / 2) = 23 / 2
Add the integrals together: 4 + 23 / 2 = 8 + 23 / 2 = 31 / 2
So, the value of the definite integral ∫(3  x  3) dx from 1 to 4 is 31 / 2 or 15.5.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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