CUET Preparation Today
The area in the first quadrant bounded by y = 4x2, x = 0 y = 1 and y = 4 is : |
2 5/2 7/3 3 |
7/3 |
Area = \(\int_{1}^{4}d\) y x = \(\int_{1}^{4}dy \frac{\sqrt{y}}{2}\) = \(\frac{1}{2} |\frac{2}{3} \sqrt{y^3}|_1^4\) = \(\frac{7}{3}\) |
