Integrals — Class 12 Maths Solution

$\int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} = \left[ {4 \cdot \cfrac{{{x^4}}}{4} - 5 \cdot \cfrac{{{x^3}}}{4} + 6 \cdot \cfrac{{{x^2}}}{2} + 9x} \right]_1^2$

$= \left[ {{x^4} - \cfrac{5}{3}{x^3} + 3{x^2} + 9x} \right]_1^2$

$= \left( {{2^4} - {1^4}} \right) - \cfrac{5}{3}\left( {{2^3} - {1^3}} \right) + 3\left( {{2^2} - {1^2}} \right) + 9\left( {2 - 1} \right)$

$= \left( {16 - 1} \right) - \cfrac{5}{3}\left( {8 - 1} \right) + 3\left( {4 - 1} \right) + 9\left( 1 \right) = 15 - \cfrac{{35}}{3} + 9 + 9$

$= 33 - \cfrac{{35}}{3} = \cfrac{{99 - 35}}{3} = \cfrac{{64}}{3}$