Question

If ${\int }_1^a(3x^2+2x+1)dx=11$, find the real values of $a$.

Answer 1

${\int }_1^a(3x^2+2x+1)dx=11$

$\Rightarrow {[\frac{3x^3}{3}+\frac{2x^2}{2}+x]}_1^a=11$

$\Rightarrow {[x^3+x^2+x]}_1^a=11$

$\Rightarrow [a^3-1+a^2-1+a-1]=11$

$\Rightarrow a^3+a^2+a-3=11$

$\Rightarrow a^3+a^2+a-14=0$

Put $a=2\Rightarrow 2^3+2^2+2-14=8+4+2-14=0$

Thus, one root is $a=2$

Again,When $a^3+a^2+a-14=0$ is divided by $a-2$ we get quotient$=a^2+3a+7$

$(a-2)(a^2+3a+7)=0$

$\Rightarrow a-2=0,a^2+3a+7=0$

$\Rightarrow a=2,\frac{-3\pm \sqrt{9-28}}{2}$

$\Rightarrow a=2,\frac{-3\pm \sqrt{-19}}{2}\notin R$

Thus,$a=2$ is the only real root.