If $\int_{1}^{k}\left(3 x^{2}+2 x+1\right) d x=11$, then $\mathrm{k}=$

Question

Aicharya · Accepted Answer

We have $\left[3\left(\frac{x^{3}}{3}ight)+2\left(\frac{x^{2}}{2}ight)+xight]_{1}^{k}=11$ $	herefore\left[x^{3}+x^{2}+xight]_{1}^{k}=11$ $\left(k^{3}+k^{2}+kight)-(1+1+1)=11 \quad \Rightarrow k^{3}+k^{2}+k=14$ $	herefore \mathrm{k}\left(\mathrm{k}^{2}+\mathrm{k}+1ight)=2 	imes 7 \Rightarrow \mathrm{k}=2$

If $\int_{1}^{k}\left(3 x^{2}+2 x+1\right) d x=11$, then $\mathrm{k}=$

Solution