# Evaluate the following integral. \int_{0}^{1}t^{\frac{5}{2}}(\sqrt{t}-3t)dt

Evaluate the following integral.
$$\int_{0}^{1}t^{\frac{5}{2}}(\sqrt{t}-3t)dt$$

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Step 1
Given:
$$\int_{0}^{1}t^{\frac{5}{2}}(\sqrt{t}-3t)dt$$
Step 2
$$\int_{0}^{1}t^{\frac{5}{2}}(\sqrt{t}-3t)dt=\int_{0}^{1}(t^{3}-3t^{\frac{7}{2}})dt$$
$$=[\frac{t^{4}}{4}-6\frac{t^{\frac{9}{2}}}{9}]_{0}^{1}$$
$$=[\frac{t^{4}}{4}-\frac{2t^{\frac{9}{2}}}{3}]_{0}^{1}$$
$$=[\frac{(1)^{4}}{4}-\frac{2(1)^{\frac{9}{2}}}{3}]-[\frac{(0)^{4}}{4}-\frac{2(0)^{\frac{9}{2}}}{3}]$$
$$=\frac{1}{4}-\frac{2}{3}-(0)$$
$$=\frac{-5}{12}$$