Question

Evaluate the definite integral: $∫^2_1(4x^3-5x^2+6x+9)dx$

Answer 1

$Let I=∫^2_1(4x^3-5x^2+6x+9)dx$

$∫(4x^3-5x^2+6x+9)dx=4(\frac{x4}{4})-5(\frac{x^3}{3})+6(\frac{x^2}{2})+9(x)$

$=x^4-\frac{5x^3}{3}+3x^2+9x=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(2)-F(1)$

$I={2^4-\frac{5.(2)^3}{3}+3(2)^2+9(2)}-{(1)^4-\frac{5(1)^3}{3}+3(1)^2+9(1)}$

$=(16-\frac{40}{3}+12+18)-(1+\frac{5}{3}-3+9)$

$=16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9$

$=33-\frac{35}{3}$

$=\frac{99-35}{3}$

$=\frac{64}{3}$