Question

The value of $\underset{1}{\overset{2}{\int }}(4x^3-5x^2+6x+9)dx$ is:

• A. $\frac{64}{3}$
• B. $\frac{8}{5}$
• C. $\frac{9}{6}$
• D. $\frac{55}{6}$

Answer 1

The correct option is A $\frac{64}{3}$

Let $I=\underset{1}{\overset{2}{\int }}(4x^3-5x^2+6x+9)dx$
Splitting the integrals, we have
$I=\underset{1}{\overset{2}{\int }}4x^{3}dx-\underset{1}{\overset{2}{\int }}5x^{2}dx+\underset{1}{\overset{2}{\int }}6xdx+\underset{1}{\overset{2}{\int }}9dx$
$=4\underset{1}{\overset{2}{\int }}{x}^{3}dx-5\underset{1}{\overset{2}{\int }}{x}^{2}dx+6\underset{1}{\overset{2}{\int }}xdx+9\underset{1}{\overset{2}{\int }}dx$
$=4{\left[\frac{x^4}{4}\right]}_1^2-5{\left[\frac{x^3}{3}\right]}_1^2+6{\left[\frac{x^2}{2}\right]}_1^2+9{\left[x\right]}_1^2$
$=[2^4-1^4]-5[\frac{2^3}{3}-\frac{1^3}{3}]+6[\frac{2^2}{2}-\frac{1^2}{2}]+9[2-1]$
$=(16-1)-5\left(\frac{7}{3}\right)+3\left(3\right)+9$
$=33-\frac{35}{3}$
$=\frac{99-35}{3}=\frac{64}{3}$