Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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9514 1404 393
Answer:
- 113.04, 4, 452.16
- 78.5, 11, 863.5
- 153.86, 10, 1538.6
Step-by-step explanation:
When calculations are repetitive, I like to let a calculator or spreadsheet do them. Here we have used the formula for the base area:
B = πr² = 3.14×r²
In the second figure, the radius is half the diameter, so is 5 units.
The table in the attachment lists the base area B, the height h (from the figures), and the volume V. These are the values that you need to drag and drop to the boxes in your problem.
V = 113.04 · 4
V = 452.16 units³ . . . showing how the numbers are used in the first figure
Answer: (-1/5, 2/5)
Step-by-step explanation:
See the photo for explanation and work.
The last one if I'm not mistaking