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$$\int_{0}^{9} f(x) \mathrm{d} x=37 \quad, \quad \int_{0}^{9} g(x) \mathrm{d} x=16 \quad \text { Find } \int_{0}^{9}(2 f(x)+3 g(x)) \mathrm{d} x$$

$$\int_{0}^{9}(2 f(x)+3 g(x)) d x=\int_{0}^{9} 2 f(x) d x+\int_{0}^{4} 3 g(x) d x$$

$$=2 \int_{0}^{9} f(x) d x+3 \int_{0}^{9} g(x) d x$$


Use the properties of the integral to verify the Inequality: $$-\frac{1}{3} \leq \int_{0}^{1} x^{2} \cos x \mathrm{d} x \leq \frac{1}{3}$$

$$-1 \leq \cos x \leq 1 \quad \quad x^{2} \geq 0$$

$$-x^{2} \leq x^{2} \cos x \leq x^{2}$$

$$-\int_{0}^{1} x^{2} d x \leq \int_{0}^{1} x^{2} \cos x d x \leq \int_{0}^{1} x^{2} d x$$

$$-\left[\frac{x^{3}}{3}\right]_{0}^{1} \leq \int_{0}^{1} x^{2} \cos x d x \leq\left[\frac{x^{3}}{3}\right]_{0}^{1}$$

$$-\left(\frac{1}{3}-0\right) \leq \int_{0}^{1} x^{2} \cos x d x \leq \frac{1}{3}-0$$

$$-1 / 3 \leq \int_{0}^{1} x^{2} \cos x d x \leq \frac{1}{3}$$

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