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Expand the integral $\int_{-1}^{4}\left(x^6+\frac{1}{3}x^2+8x-3\right)dx$ into $4$ integrals using the sum rule for integrals, to then solve each integral separately
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$\int_{-1}^{4} x^6dx+\int_{-1}^{4}\frac{1}{3}x^2dx+\int_{-1}^{4}8xdx+\int_{-1}^{4}-3dx$
Learn how to solve problems step by step online. Integrate the function x^6+1/3x^28x+-3 from -1 to 4. Expand the integral \int_{-1}^{4}\left(x^6+\frac{1}{3}x^2+8x-3\right)dx into 4 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{-1}^{4} x^6dx results in: 2340.7142857. The integral \int_{-1}^{4}\frac{1}{3}x^2dx results in: \frac{65}{9}. The integral \int_{-1}^{4}8xdx results in: 60.