CUET Preparation Today
The value of a for which $\int\limits^{a}_{1}x^2\, dx =\frac{7}{3}$ is : |
-1 1 2 -2 |
2 |
The correct answer is Option (3) → 2 $\int\limits^{a}_{1}x^2\, dx =\frac{7}{3}⇒\frac{1}{3}\left[\frac{x^3}{1}\right]^{a}_{1}=\frac{7}{3}$ so $a^3-1=7⇒a^3=8$ $a=2$ |
