# How do you evaluate the integral #int 1/sqrtxdx# from 0 to 1?

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To evaluate the integral ∫(1/√x) dx from 0 to 1, we can use the technique of substitution. Let u = √x, then du/dx = 1/(2√x), or du = (1/(2√x)) dx. This implies that dx = 2√x du.

When x = 0, u = √0 = 0, and when x = 1, u = √1 = 1. Substituting these limits and the expression for dx into the integral, we get:

∫(1/√x) dx = ∫(1/u) * 2√x du

= 2∫du from 0 to 1

= 2(u)| from 0 to 1

= 2(1) - 2(0)

= 2.

Therefore, the value of the integral ∫(1/√x) dx from 0 to 1 is 2.

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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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