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worty [1.4K]
3 years ago
13

If the integral of f(x)dx = 2.3 from 1 to 3 and F'(x) = f(x), what is the value of F(3) - F(0). The graph will be inserted as a

picture. ...?
Mathematics
2 answers:
solniwko [45]3 years ago
7 0
I think you forgot to add the attachment. But I did some research. The correct answer to this question is 2.3 + 1 = 3.3. To solve this, If f(x)dx = 2.3 and <span>F' (x) = f(x), 
</span><span>
F(3) − F(0) = 30f(x)dx 
</span>2.3 + 1 = 3.3. This <span>is the value of F(3) - F(0).
</span><span>
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 

</span>
lana66690 [7]3 years ago
4 0
<span>F(3)−F(0)=<span>∫30</span>f(x)dx</span><span>
so it looks like 2.3+1=3.3</span>
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     df =  \frac{ [ \frac{s_1^2 }{n_1 }  + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2  }

=>  df =  \frac{ [ \frac{5.9^2 }{34 }  + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2  }

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=>  SE = 1.227

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