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Zanzabum
3 years ago
10

Find the area under the curve y = 13/x3 from x = 1 to x = t. Evaluate the area under the curve for t = 10, t = 100, and t = 1000

. t = 10 t = 100 t = 1000 Find the total area under this curve for x ≥ 1.
Mathematics
1 answer:
zvonat [6]3 years ago
8 0

Answer:

t = 10

A = 32496.75

t = 100

A = 324999996.8

t = 1000

A=3.25\times 10^{12}

Step-by-step explanation:

The area under the curve is calculated by using the following definite integral:

A = \int\limits^t_ {1} \,{13\cdot x^{3}}  dx

A = 13 \int\limits^t_1 {x^{3}} \, dx

A = \frac{13}{4}\cdot (t^{4}-1)

Evaluated areas are presented below:

t = 10

A = 32496.75

t = 100

A = 324999996.8

t = 1000

A=3.25\times 10^{12}

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\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}

Cancelling the common terms, we get

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)

Substituting we get,

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}

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