Hello fellows,
Just like what I did in my derivative question, I'd also like to ask if my answer is correct in integrating.
Given of
\(\int_2^3 2x^{3}+11x^{2}+31x.dx\)
\(=\int 2x^{3}+\int 11x^{2}+\int 31x\)
\(=\frac {2x^{3+1}}{3+1}+\frac{11x^{2+1}}{2+1}+\frac{31x^{1+1}}{1+1}\)
\(=\frac {2x^{4}}{4}+\frac{11x^{3}}{3}+\frac{31x^{2}}{2}\)
\(=\frac {x^{4}}{2}+\frac{11x^{3}}{3}+\frac{31x^{2}}{2}\)
\(=\frac {3x^{4}+22x^{3}+93x^{2}}{6}\)\(\]_2^3\)
At x = 2
\(=\frac {3(2^{4})+22(2^{3})+93(2^{2})}{6}\)
\(=\frac {3(16)+22(8)+93(4)}{6}\)
\(=\frac {48+176+372}{6}\)
\(=\frac {576}{6}\)
At x = 3
\(=\frac {3(3^{4})+22(3^{3})+93(3^{2}}{6}\)
\(=\frac {3(81)+22(27)+93(9)}{6}\)
\(=\frac {243+594+837}{6}\)
\(=\frac {1674}{6}\)
Subtracting:
\(\frac {1674}{6}-\frac {596}{6}\)
\(=\frac {1078}{6}\)
\(=179.\overline{6}\)
Thank you very much!
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