Question

Guys pls ı need help really....

Answer 1

Answer:

Step-by-step explanation:

Calculus....finally something FUN to answer!

This is integration, but here you have the area under the curve, you just don't have the upper bound. Setting that problem up looks like this:

$28=\int\limits^b_1 {7x} \, dx$  and using the First Fundamental Theorem of Calculus, that will integrate to look like this:

$28=\frac{7x^2}{2}$ from 1 to b and applying the FTC:

$28=\frac{7b^2}{2}-\frac{7}{2}$  Multiplying everything by 2 to get rid of the denominators gives you:

$56=7b^2-7\\\\56=7(b^2-1)\\8=b^2-1\\9=b^2\\b=3$