Question

(Can someone check my work please).
Use the form of the definition of the integral given in the theorem to evaluate the integral. Question #8, picture included.

Answer 1

\[\lim_{n \rightarrow \infty} 6/n[(1+6i/n)^2-4(1+6i/n)+5]\]
\[\lim_{n \rightarrow \infty} 6/n[(1+12i/n)+(36i^2/n^2)-4-(24i/n)+5]\]
\[\lim_{n \rightarrow \infty} 6/n[(36i^2/n^2)-(12i/n)+2]\]
\[\lim_{n \rightarrow \infty} [(216i^2/n^3)-(72i/n^2)+2]\]
\[\lim_{n \rightarrow \infty} [(216/n^3)*i^2-(72/n^2)*i+2] \]
\[\lim_{n \rightarrow \infty} [(216/n^3)*\frac{ n(n+1)(2n+1 }{ 6 }-(72/n^2)*\frac{ n(n+1) }{ 2 }+2(n)*\frac{ 6 }{ n }]\]
\[\lim_{n \rightarrow \infty} [(216/6) * \frac{ n+1 }{ n }*\frac{ 2n+1 }{ n }-(72/2)*\frac{ n+1 }{ n }+12]\]
\[\lim_{n \rightarrow \infty} [36(1+\frac{ 1 }{ n })(2+\frac{ 1 }{ n }) - 36(1+\frac{ 1 }{ n })+12]\]
\[= 36(1)(2)-36(1)+12 = 48\]

Is 48 the correct answer?

Answer 2

bingo!!

https://www.wolframalpha.com/input/?i=integrate+x%5E2+-+4x+%2B+5+dx+from+x+%3D+1+to+x+%3D+7

Answer 3

@IrishBoy123 Thanks, I need to use wolfram to check my answers more often. :D

Answer 4

sounds like a great idea. i will try that too!