So, the definite integral 
Given that
We find

<h3>Definite integrals </h3>
Definite integrals are integral values that are obtained by integrating a function between two values.
So, 
So, ![\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E1_0%20%7B%284%20-%206x%5E%7B2%7D%20%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B4%7D%20%5C%2C%20dx%20-%20%5Cint%5Climits%5E1_0%20%7B6x%5E%7B2%7D%20%7D%20%5C%2C%20dx%20%5C%5C%3D%20%204%5Bx%5D%5E%7B1%7D_%7B0%7D%20%20%20%20-%20%5Cint%5Climits%5E1_0%20%7B6x%5E%7B2%7D%20%7D%20%5C%2C%20dx%20%5C%5C%3D%20%204%5Bx%5D%5E%7B1%7D_%7B0%7D%20%20%20%20-%206%5Cint%5Climits%5E1_0%20%7Bx%5E%7B2%7D%20%7D%20%5C%2C%20dx%20%5C%5C%3D%204%5B1%20-%200%5D%20%20%20%20-%206%5Cint%5Climits%5E1_0%20%7Bx%5E%7B2%7D%20%7D%20%5C%2C%20dx%5C%5C%3D%204%5B1%5D%20%20%20%20-%206%5Cint%5Climits%5E1_0%20%7Bx%5E%7B2%7D%20%7D%20%5C%2C%20dx%5C%5C%3D%204%20%20%20%20-%206%5Cint%5Climits%5E1_0%20%7Bx%5E%7B2%7D%20%7D%20%5C%2C%20dx)
Since
,
Substituting this into the equation the equation, we have

So, 
Learn more about definite integrals here:
brainly.com/question/17074932
Answer:
98 sweets
Step-by-step explanation:
Caroline got 14 sweets, ratio for her was 1.
Krutika had twice the bigger ratio than her, which means you have to multipy Carolines' sweets by 2. -> 14*2= 28
And do the same thing for Natasha which gives us the result of 56. 14*4= 56
And finally, you add all of those together and get 14+28+56=98.
Hope I helped! :)
If this paper dosent help u solve it on Your own i can walk u through it
Answer:
Im pretty sure its the first choice but the question is hard to read
Step-by-step explanation:



To multiply powers of the same base, add their exponents.

Add 5 and 2 to get 7.

- The first option <em><u>n⁷</u></em><em><u> </u></em>is the correct answer.