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5. [2/4 Points] DETAILS PREVIOUS ANSWERS LARCALCET7 5.2.053. Use the limit process to find the area of the region between the

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\text{ Area bounded by the graph } y= f(x) \text{ and x-axis over the interval [a, b] is given by}

\text{ Area } = \int_a^b f(x)dx =\lim_{n\to\infty} \sum_{i=1}^nf(x_i) \bigtriangleup x

\text{ Given }

y 1 = 64 -

2.4

\text{ We have, }a =2,\:\:\: b=4,\:\:\:\: f(x) =64-x^3

\text{Let n be number of sub intervals of [2, 4]}

\text{Width of each sub intervals, } \bigtriangleup x =\frac{b-a}{n}=\frac{4-2}{n}=\frac{2}{n}

x_i =a+i\bigtriangleup x =2+i\left ( \frac{2}{n} \right )=2+\frac{2i}{n}

\text{Calculating area using limit process}

\text{ Area } = \int_a^b f(x)dx =\lim_{n\to\infty} \sum_{i=1}^nf(x_i) \bigtriangleup x

=> \text{ Area } =\lim_{n\to\infty} \sum_{i=1}^n (64-x_i^3) \left ( \frac{2}{n} \right )=\lim_{n\to\infty} \sum_{i=1}^n \left ( 64-\left ( 2+\frac{2i}{n} \right )^3 \right ) \left ( \frac{2}{n} \right )

=> \text{ Area } =\lim_{n\to\infty} \sum_{i=1}^n \left ( 64-\left ( 8+\frac{24i}{n}+\frac{24i^2}{n^2} +\frac{8i^3}{n^3}\right ) \right ) \left ( \frac{2}{n} \right )=\lim_{n\to\infty} \sum_{i=1}^n \left ( 56-\frac{24i}{n}-\frac{24i^2}{n^2} -\frac{8i^3}{n^3} \right ) \left ( \frac{2}{n} \right )

=> \text{ Area } =\lim_{n\to\infty} \sum_{i=1}^n \left ( \frac{112}{n}-\frac{48i}{n^2}-\frac{48i^2}{n^3} -\frac{16i^3}{n^4} \right )

=> \text{ Area } =\lim_{n\to\infty} \left ( \sum_{i=1}^n \frac{112}{n} -\sum_{i=1}^n \frac{48i}{n^2}-\sum_{i=1}^n \frac{48i^2}{n^3} -\sum_{i=1}^n \frac{16i^3}{n^4} \right )

=> \text{ Area } =\lim_{n\to\infty} \left (\frac{112}{n} \sum_{i=1}^n(1) - \frac{48}{n^2} \sum_{i=1}^n(i)-\frac{48}{n^3}\sum_{i=1}^ni^2 - \frac{16}{n^4}\sum_{i=1}^ni^3 \right )

=> \text{ Area } =\lim_{n\to\infty} \left (\frac{112}{n} (n) - \frac{48}{n^2} \left (\frac{n(n+1)}{2} \right )-\frac{48}{n^3} \left (\frac{n(n+1)(2n+1)}{6} \right ) - \frac{16}{n^4} \left (\frac{n(n+1)}{2} \right )^2 \right )

=> \text{ Area } =\lim_{n\to\infty} \left (112 - \frac{48}{n^2} \left (\frac{n^2 \left ( 1+\frac{1}{n} \right )}{2} \right )-\frac{48}{n^3} \left (\frac{n^3\left ( 1+\frac{1}{n} \right )\left ( 2+\frac{1}{n} \right )}{6} \right ) - \frac{16}{n^4} \left (\frac{n^2\left ( 1+\frac{1}{n} \right )}{2} \right )^2 \right )

=> \text{ Area } =\lim_{n\to\infty} \left (112 - 24\left ( 1+\frac{1}{n} \right ) -8\left ( 1+\frac{1}{n} \right )\left ( 2+\frac{1}{n} \right ) -4 \left ( 1+\frac{1}{n}\right )^2 \right )

=> \text{ Area } = \left (112 - 24\left ( 1+0 \right ) -8\left ( 1+0 \right )\left ( 2+0 \right ) -4 \left ( 1+0\right )^2 \right )

=> \text{ Area } =68

Hence,

\text{ Area } =68

500 y 400 300 200 100 X o -10 -5 10 -100 -200 -300 -400

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