Answer:
6 < 63, TRUE
Step-by-step explanation:
[tex]\sf \int\limits^4_2 {x} \, dx < \int\limits^6_3 {x^2} \, dx[/tex]
[tex]\implies \sf \big[\frac{x^{1+1}}{1+1}\big]_{2}^{4} < \big[\frac{x^{2+1}}{2+1}\big]_{3}^{6}[/tex]
[tex]\implies \sf \big[\frac{x^2}{2}\big]_{2}^{4}<\big[\frac{x^3}{3}\big]_{3}^{6}[/tex]
[tex]\implies \sf \big(\frac{4^2}{2} -\frac{2^2}{2}\big) < \big(\frac{6^3}{3}-\frac{3^3}{3}\big)[/tex]
[tex]\implies \sf \big(\frac{16}{2}-\frac{4}{2}\big)<\big(\frac{216}{3}-\frac{27}{3}\big)[/tex]
[tex]\implies \sf (8 - 2) < (72-9)[/tex]
[tex]\implies \sf 6 < 63[/tex]
Hence, the inequality is TRUE.
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