# How do you find the definite integral for: #[(6x^2+2) / sqrt(x)] dx # for the intervals #[1, 5]#?

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To find the definite integral of ( \frac{6x^2 + 2}{\sqrt{x}} ) over the interval ([1, 5]), you can use the Fundamental Theorem of Calculus. First, rewrite the expression as ( 6x^2x^{-1/2} + 2x^{-1/2} ). Then integrate each term separately:

[ \int_{1}^{5} \frac{6x^2 + 2}{\sqrt{x}} , dx = \int_{1}^{5} 6x^{2}x^{-1/2} , dx + \int_{1}^{5} 2x^{-1/2} , dx ]

[ = \left[ 6 \cdot \frac{x^{2 + 1}}{2 + 1} \right]*{1}^{5} + \left[ 2 \cdot \frac{x^{1/2 + 1}}{1/2 + 1} \right]*{1}^{5} ]

[ = \left[ 2x^{5/2} \right]*{1}^{5} + \left[ 4x^{3/2} \right]*{1}^{5} ]

[ = 2(5^{5/2} - 1) + 4(5^{3/2} - 1) ]

[ \approx 101.656 ]

Therefore, the definite integral of ( \frac{6x^2 + 2}{\sqrt{x}} ) over the interval ([1, 5]) is approximately ( 101.656 ).

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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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