Question

@perl calc help!

Answer 1

@perl please help!

Answer 2

@campbell_st please help!

Answer 3

so if we let a= 4, and delta x = (b-a)/n = 3/n
we get

(b - 4) /n = 3/n

b-4 = 3

b = 7

Answer 4

any ideas?

Answer 5

$$

\Large{ \int_{a}^{b} f(x) dx=\lim_{n\to \infty} \sum_{k=1}^{n}f(a + k\cdot \Delta x )\cdot \Delta x
\\ where ~~\Delta x = (b-a)/n
\\ \therefore \\
\lim_{n\to \infty} \sum_{k=1}^{n}f(4 + k\cdot 3/n )\cdot 3/n = \int_{4}^{7} x^2 dx

}

$$