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# (a) Graph the curve $y = x \sqrt[3]{4 - x}$ , $0 \le x \le 4$.(b) Compute the lengths of inscribed polygons with $n = 1$, 2, and 4 sides. (Divide the interval into equal sub-intervals.) Illustrate by sketching these polygons (as in Figure 6).(c) Set up an integral for the length of the curve.(d) Use your calculator to find the length of the curve to four decimal places. Compare with the approximations in part (b).

## a) Let $f(x)=y=x \sqrt[3]{4-x}$ with $0 \leq x \leq 4$b) $\sqrt{1+(\sqrt[3]{3})^{2}}+\sqrt{\left(1+(2 \sqrt[3]{2}-\sqrt[3]{3})^{2}\right.}+\sqrt{1+(3-2 \sqrt[3]{2})^{2}}+\sqrt{1+9} \approx 7.5$c) $L=\int_{0}^{4} \sqrt{1+\left[\frac{12-4 x}{3(4-x)^{2 / 3}}\right]^{2}} d x$d) $L \approx 7.79879$

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Applications of Integration

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### Video Transcript

he has clear. So when you read here, so for part, they already drew the graph. So it looks something like this to curve for part B. We're gonna find the link, so you know that the bottom is for so from zero half of zero and four comma effort for building this For now, if we look it this polygon from zero comma zero, 22 comma F comma too. We see that feeling is going to be so from 00 comma zero two comma F comma till it gives us two comma too Que brute go to and of for comma zero. So the length is approximately 6.43 And finally the length for the 0.0 comma zero one comma Q Bert of three to Comack, You brute Too common to cube root of two three com a three and four comma zero The length is approximately 7.5 for part C. We're gonna find the arc length equation So we have our why is equal to x cuba it of four minus x. So are derivative. Gives us 12 minus four x over three arms form minus X to the 2/3 power. So l is equal to from 0 to 4 square root of one plus 12 minus for X over three times four minus X to the 2/3 power. And we square this the ex finally, for party. We're just gonna need a calculator and we get approximately 7.79 879 here, in part being we found the length of the curve using polygons, and we see that these are all under estimates. So we conclude that the actual value is larger since for our approximations, we're using it were using straight lines.

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Applications of Integration

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