Find the area of the region described. The region in the first quadrant bounded by \( \grave{y}=x^{\frac{2}{3}} \) and \( y=5 \) The area of the region is \( \square \). (Type an exact answer.)

To find the area of the region bounded by the curve \( y = x^{\frac{2}{3}} \) and the line \( y = 5 \) in the first quadrant, we first need to determine the points of intersection. Setting \( x^{\frac{2}{3}} = 5 \), we have: \[ x^{\frac{2}{3}} = 5 \implies x = 5^{\frac{3}{2}} = 5 \sqrt{5} = 5^{1.5} = \frac{25}{5^\frac{1}{2}} = 25 \div \sqrt{5} \] Next, we set up the integral to find the area. The area \( A \) can be calculated by integrating from \( x = 0 \) to \( x = 25 \div \sqrt{5} \): \[ A = \int_0^{25 \div \sqrt{5}} (5 - x^{\frac{2}{3}}) \, dx \] Let's compute this integral step-by-step: 1. We calculate the integral of \( 5 \): \[ \int 5 \, dx = 5x \] 2. We calculate the integral of \( x^{\frac{2}{3}} \): \[ \int x^{\frac{2}{3}} \, dx = \frac{3}{5} x^{\frac{5}{3}} \] Putting it together, the area becomes: \[ A = \left[ 5x - \frac{3}{5} x^{\frac{5}{3}} \right]_0^{25 \div \sqrt{5}} \] Evaluating at the limits: For \( x = 25 \div \sqrt{5} \): \[ A = 5 \left( \frac{25}{\sqrt{5}} \right) - \frac{3}{5} \left( \frac{25}{\sqrt{5}} \right)^{\frac{5}{3}} \] Calculate \( 5 \left( \frac{25}{\sqrt{5}} \right) = 5 \cdot 5^{2} = 125 \). For \( x = 25 \div \sqrt{5} \): Calculating \( \left( \frac{25}{\sqrt{5}} \right)^{\frac{5}{3}} = \frac{25^{\frac{5}{3}}}{5^{\frac{5}{6}}} = 25^{\frac{5}{3}} \left( 5^{- \frac{5}{6}} \right)\): \[ 25^{\frac{5}{3}} = (5^2)^{\frac{5}{3}} = 5^{\frac{10}{3}} \] Thus: \[ A = 125 - \frac{3}{5} (25 \cdot 5^{2})^{\frac{5}{3}} = 125 - \frac{3}{5} 125(5^{- \frac{1}{3}}) = 125 - 75 = 50 \] Finally, the area of the region is \( \boxed{50} \).