Find the area between the curve and the $$ x $$-axis over the indicated interval. $$ y=25-x^{2} ;[-5,5] $$ The area under the curve is $$ \square $$. (Simplify your answer.)

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The area is $$ \frac{500}{3} $$ square units. Find the integral $$ \int_{-5}^{5} (25-x^{2}) \,dx $$. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: $$\int_{-5}^{5} 25-x^{2} dx$$ - step1: Evaluate the integral: $$\int 25-x^{2} dx$$ - step2: Use properties of integrals: $$\int 25 dx-\int x^{2} dx$$ - step3: Evaluate the integral: $$25x-\frac{1}{3}x^{3}$$ - step4: Return the limits: $$\left(25x-\frac{1}{3}x^{3}\right)\bigg |_{-5}^{5}$$ - step5: Calculate the value: $$\frac{500}{3}$$ The area between the curve $$ y=25-x^{2} $$ and the $$ x $$-axis over the interval $$[-5,5]$$ is $$ \frac{500}{3} $$ or $$ 166\frac{2}{3} $$ or $$ 166.6 $$ square units.

Find the area between the curve and the \( x \)-axis over the indicated interval. \( y=25-x^{2} ;[-5,5] \) The area under the curve is \( \square \). (Simplify your answer.)

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Find the area between the curve and the \( x \)-axis over the indicated interval. \( y=25-x^{2} ;[-5,5] \) The area under the curve is \( \square \). (Simplify your answer.)

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