Question

If ${\displaystyle {\int }_{-2}^3{x}^{2}dx=k{\int }_0^2{x}^{2}dx+{\int }_2^3{x}^{2}dx}$, then the value of k is ______.

पर्याय

• 2

• 1

• 0

• ${\displaystyle \frac{1}{2}}$

Answer 1

If ${\displaystyle {\int }_{-2}^3{x}^{2}dx=k{\int }_0^2{x}^{2}dx+{\int }_2^3{x}^{2}dx}$, then the value of k is 2.

Explanation:

${\displaystyle {\int }_{-2}^3{x}^{2}dx=k{\int }_0^2{x}^{2}dx+{\int }_2^3{x}^{2}dx}$

⇒ ${\displaystyle {\left[\frac{x^3}{3}\right]}_{-2}^3=k{\left[\frac{x^3}{3}\right]}_0^2+{\left[\frac{x^3}{3}\right]}_2^3}$

⇒ ${\displaystyle [9+\frac{8}{3}]=k\left[\frac{8}{3}\right]+[9-\frac{8}{3}]}$

⇒ k = 2