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Afina-wow [57]
3 years ago
10

Rob and his two friends were digging holes to plant some new trees. The holes need to be at least 3 feet deep. So far robs hole

is 14/5, sams is 2.6 feet and toms is 2 1/8 feet. Who is closest to being finished
Mathematics
1 answer:
Greeley [361]3 years ago
7 0

Rewrite all the numbers as decimals:

14/5 = 2.8

2.6 is already a decimal

2 1/8 = 2.125

2.8 is the largest number so rob is closest to being finished.

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0
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So, the definite integral  \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

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\int\limits^1_0 {x^{2} } \, dx = 13

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\int\limits^1_0 {(4 - 6x^{2} )} \, dx

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4    - 6\int\limits^1_0 {x^{2} } \, dx

Since

\int\limits^1_0 {x^{2} } \, dx = 13,

Substituting this into the equation the equation, we have

\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Learn more about definite integrals here:

brainly.com/question/17074932

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