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alina1380 [7]
3 years ago
7

Help me with this question, please, ill choose brainiest and give u 30 points for answering

Mathematics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer: D

Step-by-step explanation:

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Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

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3 years ago
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DaniilM [7]

Answer:

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Step-by-step explanation:

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Choice D is not right because it was 5 games because there is 5 dots above 3

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