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Hunter-Best [27]
2 years ago
15

1. Find the area of the region bounded above by y = e^x, bounded below by y = x, and bounded on the sides

Mathematics
1 answer:
salantis [7]2 years ago
3 0

The area of the region bounded above by y= eˣ bounded by y = x, and bounded on the sides; x =0; and x = 1 is given as e¹ - 1.5.

<h3>What is the significance of "Area under the curve"?</h3>

This is the condition in which one process increases a quantity at a certain rate and another process decreases the same quantity at the same rate, and the "area" (actually the integral of the difference between those two rates integrated over a given period of time) is the accumulated effect of those two processes.

<h3>What is the justification for the above answer?</h3>

Area = \int\limits^1_0 {(e^x-x)} \, dx

= \int\limits^1_0 {e^x-x} \, dx - \int\limits^1_0 {(x)} \, dx

= e¹-(1/2-0); or

Area = e -1.5 Squared Unit

The related Graph is attached accordingly.

Learn more about area bounded by curve:
brainly.com/question/27866606
#SPJ1

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One night Puvi spent 1/3 of his money on dinner and then 1/4 of the remaining money on dessert. He then had just enough left so
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Answer:

Step-by-step explanation:

Cost of two tickets = 2 * 30 = £60

Let the amount with Puvi = £x

Money spent on dinner = (1/3)*x

                                        = \dfrac{1}{3}x

Remaining \ money =x -  \dfrac{1}{3}x =\dfrac{3}{3}x-\dfrac{1}{3}x=\dfrac{2}{3}x

Money \ spent \ on \ dessert = \dfrac{1}{4} *\dfrac{2}{3}x =\dfrac{1}{6}x\\\\\\

Total money - money spent on dinner - money spent on dessert = 60

x -\dfrac{1}{3}x -\dfrac{1}{6}x=60\\\\\\\dfrac{6}{6}x-\dfrac{1*2}{3*2}-\dfrac{1}{6}x =60\\\\\\\dfrac{6}{6}x-\dfrac{2}{6}x-\dfrac{1}{6}x=60\\\\\\\dfrac{6-2-1}{12}x=60\\\\\\\dfrac{3}{12}x=60\\\\\\x=60*\dfrac{12}{3}=60*4=240

Money that Puvi had at the start of the night = £ 240

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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