By using the properties of definite integrals | vedclass

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(A) Let $I = \int_{0}^{2} x \sqrt{2-x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{2} (2-x

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$16\sqrt{2}/15$

Vedclass · Answer

(A) Let $I = \int_{0}^{2} x \sqrt{2-x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{2} (2-x) \sqrt{2-(2-x)} \, dx = \int_{0}^{2} (2-x) \sqrt{x} \, dx$. $I = \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx$. Integrating term by term: $I = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_{0}^{2} = \left[ \frac{4}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{0}^{2}$. Substituting the limits: $I = \left( \frac{4}{3} (2)^{3/2} - \frac{2}{5} (2)^{5/2} \right) - 0$. Since $(2)^{3/2} = 2\sqrt{2}$ and $(2)^{5/2} = 4\sqrt{2}$: $I = \frac{4}{3} (2\sqrt{2}) - \frac{2}{5} (4\sqrt{2}) = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}$. Taking the common denominator: $I = \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15}$.

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2} x \sqrt{2-x} \, dx$.

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