$01-917)$ Calculus I $24/25(1)$

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Question $3$

The expression $\frac{1}{4}ln(8+\frac{1}{4})+\frac{1}{4}ln(8+\frac{1}{2})+\frac{1}{4}ln(8+\frac{3}{4})+\frac{1}{4}ln(8+1)$ is a Riemann sum approximation of which of the following integrals?

Your answer:

${\displaystyle \underset{3}{\overset{4}{}}}ln(5+\frac{x}{4})dx$

$\frac{1}{4}{\displaystyle \underset{0}{\overset{4}{}}}ln(5+x)dx$

$\frac{1}{4}{\displaystyle \underset{3}{\overset{4}{}}}ln(5+x)dx$

${\displaystyle \underset{3}{\overset{4}{}}}ln(5+x)dx$

${\displaystyle \underset{0}{\overset{4}{}}}ln(3+x)dx$

Clear answer