Which of the following definite integrals is equivalent to...

Which of the following definite integrals is equivalent to the Riemann Sum $\lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n}$? $\int_{0}^{1} \sqrt{\frac{1}{x}} dx$ $\int_{0}^{1} \sqrt{x} dx$ $\int_{0}^{1} 4\sqrt{4x} dx$ $\int_{0}^{1} 4x\sqrt{4x} dx$

Transcript

00:01 Okay, we want to solve the integral of x to the n times natural log of x given that n is not equal to negative 1 and to do this we'll be using integration by parts where the given formula is we have uv minus the integral of v d u so to do this we have to select a u and then we have to select a dv from our function and then from there we can go forward with the integration by parts method so in this scenario we're gonna let we're gonna let you be equal to natural log of x and dv be equal to x to the n so this means we take our d u which is just a derivative of natural log of x which happens to be 1 over x and then we take our v which is the integral of dv and this just means we do the power rule so we add 1 to the exponent and then bring down the change exponent as a factor in the denominator which is n plus 1 so this means we have our u v and so this means we just write out what our integration by parts formula is.

01:10 So we have u times v.

01:12 So we have x to the m plus 1 or m plus 1 times natural log of x minus the integral of v to u where v is again x to the m plus 1 over m plus 1 times 1 over x d x so this means right here this first term we're done with that so that is going to remain here while we focus on and we focus on this other integral that we just got from the formula so this means we're going to simplify this integral down we'll change the color to green here so this means we now have the integral of x the m plus 1 over x which such just gives us x to n because if you're tracking the exponents and then we have still the n plus 1 in the denominator d x so this means we're only taking the integral of x to the because again, this n plus 1 factor in the denominator is just a constant.

02:22 So that can be brought out of the integral.

02:24 So this means once we solve this integral, we have the factor of 1 over m plus 1 times the interval of x to the end, which again, as we've done already, is x to the n plus 1 over another factor of m plus 1.

02:37 And of course, we have our constant plus c...

Question

Which of the following definite integrals is equivalent to the Riemann Sum $\lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n}$? $\int_{0}^{1} \sqrt{\frac{1}{x}} dx$ $\int_{0}^{1} \sqrt{x} dx$ $\int_{0}^{1} 4\sqrt{4x} dx$ $\int_{0}^{1} 4x\sqrt{4x} dx$

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