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Solve the product $-\frac{1}{8}\left(16-x^2\right)$
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$\int_{0}^{4}\left(\sqrt{16-x^2}-2+\frac{1}{8}x^2\right)dx$
Learn how to solve definite integrals problems step by step online. Integrate the function (16-x^2)^1/2-1/8(16-x^2) from 0 to 4. Solve the product -\frac{1}{8}\left(16-x^2\right). Expand the integral \int_{0}^{4}\left(\sqrt{16-x^2}-2+\frac{1}{8}x^2\right)dx into 3 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{0}^{4}\sqrt{16-x^2}dx results in: 4\pi . The integral \int_{0}^{4}-2dx results in: -8.