10 and I know this because I just looked it up hopefully you have a great rest of ur day
62=14 + 2x
62-14=2x
48=2x
48/2=x
24 answer: 24
hope this helped
Answer:

Step-by-step explanation:
The definite integral of a continuous function <em>f</em> over the interval [a,b] denoted by
, is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,

where
and 
To evaluate the integral

you must:
Find 

Find 

Therefore,


![\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%207%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B7%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%207%5B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29)

![\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7B4i%5E2%7D%7Bn%5E2%7D-%5Cfrac%7B6i%7D%7Bn%7D%2B2%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7B4i%5E2%7D%7Bn%5E2%7D-%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B6i%7D%7Bn%7D%2B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D2%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Cfrac%7B4%7D%7Bn%5E2%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Di%5E2%20-%5Cfrac%7B6%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Di%2B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D2%5D)
We can use the facts that


![\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Cfrac%7B4%7D%7Bn%5E2%7D%5Ccdot%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D-%5Cfrac%7B6%7D%7Bn%7D%5Ccdot%20%20%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%2B2n%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B-n%2B%5Cfrac%7B2%5Cleft%28n%2B1%5Cright%29%5Cleft%282n%2B1%5Cright%29%7D%7B3n%7D-3%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%5Cleft%28n%5E2-3n%2B2%5Cright%29%7D%7B3n%5E2%7D)

Thus,

It would be -4 cuz when x = -3, y=-4