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astra-53 [7]
2 years ago
5

6. What is the minimum vertical distance between the parabolas

t%20%7B%20and%20%7D%20y%3Dx-x%5E%7B2%7D%20%3F%20" id="TexFormula1" title=" y=x^{2}+1 \text { and } y=x-x^{2} ? " alt=" y=x^{2}+1 \text { and } y=x-x^{2} ? " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Arada [10]2 years ago
5 0

Answer:

0.75.

Step-by-step explanation:

1) the required vertical distance is the distance between two vertexes.

2) according to the item 1, the coordinates of the vertexes are:

(y=x²+1): x₀=0; y₀=1;

(y=x-x²): x₀=0.5; y₀=0.25;

3) according to the item 2, the distance between 1 and 0.25 is 0.75.

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In the x direction we consider the m=2 subintervals [0, 1] and [1, 2] (each with length 1), while in the y direction we consider the n=2 subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of R are (1, 0), (2, 0), (1, 2), (2, 2).

Let f(x,y)=7x+5y^2. The volume of the solid is approximately

\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}

###

More generally, the lower-right-corner Riemann sum over m=\mu and n=\nu subintervals would be

\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)

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