Question

निम्नलिखित प्रश्न में निश्चित समाकलन का मान ज्ञात कीजिए।

${\displaystyle {\int }_1^2(4x^3-5x^2+6x+9)dx}$

Answer 1

${\displaystyle ={[\frac{4x^4}{4}-\frac{5x^3}{3}+\frac{6x^2}{2}+9x]}_1^2}$

${\displaystyle ={(x^4-\frac{5}{3} x^3 +3x^2+9x)}_1^2}$

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

${\displaystyle =(16-1)-\frac{5}{3}(8-1)+3(4-1)+9\left(1\right)}$

${\displaystyle =15-\frac{35}{3}+9+9}$

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

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

${\displaystyle =\frac{64}{3}}$