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Sagot :
The length of the function y = 3x over the given interval [0, 2] is 3.2 units
For given question,
We have been given a function y = 3x
We need to find the length of the function on the interval x = 0 to x = 2.
Let f(x) = 3x where f(x) = y
We have f'(x) = 3, so [f'(x)]² = 9.
Then the arc length is given by,
[tex]\int\limits^a_b {\sqrt{1+[f'(x)]^2} } \, dx\\\\= \int\limits^2_0 {\sqrt{1+9} }\, dx\\\\=\sqrt{10}\\\\ =3.2[/tex]
This means, the arc length is 3.2 units.
Therefore, the length of the function y = 3x over the given interval [0, 2] is 3.2 units
Learn more about the arc length here:
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