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Dahasolnce [82]
3 years ago
7

Find the edge length of the cube. Write the answer as a fraction in simplest form. volume is 1/27 ft. cubed

Mathematics
1 answer:
Slav-nsk [51]3 years ago
6 0

The volume of a cube is the edge length cubed (S^3)

To find the edge length when the volume is known find the cubic root of the volume.

Edge = ∛1/27

Edge = 1/3 ft.

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Use the form of the definition of the integral given in the theorem to evaluate the integral. 9 (x2 − 4x + 6) dx
bagirrra123 [75]

Answer:

\mathbf{\dfrac{392}{3}}

Step-by-step explanation:

Given integral :

\int ^9_1 (x^2 -4x + 6) \ dx

Let consider f(x) = x^2 - 4x + 6 .

Then , using the formula:

\int ^b_a  \ f(x)  \ dx = \lim \limits_{x \to \infty}  \ \Delta x \ \sum \limits ^{n} _{i=1}  \  f(a + i \Delta x)

where;

a = 1 ,  b = 9 and  \Delta \ x = \dfrac{b-a}{n}

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\Delta \ x = \dfrac{8}{n}

∴

\int ^9_1 (x^2 -4x + 6) \ dx = \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ f(1 + i \dfrac{8}{n})

= \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix}  (1+ i \dfrac{8}{n})^2  - 4(1 + i \dfrac{8}{n} ) +6\end {bmatrix}

= \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix}  1+ \dfrac{64}{n^2}i^2  + \dfrac{16 }{n} i - 4 - \dfrac{32}{n}i +6\end {bmatrix}

= \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2}i^2  - \dfrac{16 }{n} i +3\end {bmatrix}

= \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2}   \sum \limits ^{n}_{i=1} \ i^2 - \dfrac{16 }{n} \sum \limits ^{n}_{i=1} \ i +3 \sum \limits ^{n}_{i=1} \ 1 \end {bmatrix}

= \lim  \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2}  \dfrac{n(n+1)(2n+1)}{6}-\dfrac{16}{n} \dfrac{n(n+1)}{2} + 3n \end {bmatrix}

= \lim  \limits _{x \to \infty} \ \begin {bmatrix} \dfrac{512}{6} (1+ \dfrac{1}{n})(2 + \dfrac{1}{n}) -64 (1 + \dfrac{1}{n} ) +24n \end {bmatrix}

= \begin {bmatrix} \dfrac{512}{6} (1+ 0)(2 + 0) -64 (1 +0 ) +24 \end {bmatrix}

= \begin {bmatrix} \dfrac{512}{6} (2)-64  +24 \end {bmatrix}

= \dfrac{512}{3}-40

= \dfrac{512- 120}{3}

= \mathbf{\dfrac{392}{3}}

7 0
2 years ago
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