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Since the integral $\int_{-2}^{3}\frac{21}{x^4}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{-2}^{0}\frac{21}{x^4}dx+\int_{0}^{3}\frac{21}{x^4}dx$
Learn how to solve problems step by step online. Integrate the function 21/(x^4) from -2 to 3. Since the integral \int_{-2}^{3}\frac{21}{x^4}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{-2}^{0}\frac{21}{x^4}dx results in: 21\lim_{c\to0}\left(\frac{1}{-3c^{3}}-\frac{1}{24}\right). The integral \int_{0}^{3}\frac{21}{x^4}dx results in: 21\lim_{c\to0}\left(\frac{1}{-81}+\frac{-1}{-3c^{3}}\right). Gather the results of all integrals.
