19. \( \int_{3}^{3}\left(x^{2}+2 x-4\right) d x=\left[\frac{1}{3} x^{3}+x^{2}-4 x\right]_{1}^{3}=(9+9-12)-\left(\frac{1}{3}+1-4\right)=6+\frac{4}{3}=\frac{29}{3} \) 21. \( \int_{0}^{2}\left(\frac{4}{5} t^{3}-\frac{4}{4} t^{2}+\frac{2}{5} t\right) d t=\left[\frac{1}{3} t^{4}-\frac{1}{4} t^{3}+\frac{4}{8} t^{2}\right]_{0}^{2}=\left(\frac{4}{5}-2+\frac{4}{5}\right)-0=2 \) 23. \( \int_{1}^{3} \sqrt{x} d x=\int_{1}^{9} x^{1 / 2} d x=\left[\frac{x^{3 / 2}}{3 / 2}\right]_{1}^{3}=\frac{4}{4}\left[x^{3 / 2}\right]_{1}^{0}=\frac{4}{4}\left(9^{3 / 2}-1^{3 / 2}\right)=\frac{3}{3}(27-1)=\frac{24}{3} \) 25. \( \int_{\pi / 6}^{\pi} \sin \theta d \theta=[-\cos \theta]_{\pi / 6}^{\pi}=-\cos \pi-\left(-\cos \frac{\pi}{6}\right)=-(-1)-(-\sqrt{3} / 2)=1+\sqrt{3} / 2 \) 27. \( \int_{0}^{1}(u+2)(u-3) d u=\int_{0}^{1}\left(u^{2}-u-6\right) d u=\left[\frac{1}{3} u^{3}-\frac{1}{2} u^{2}-6 u\right]_{0}^{1}=\left(\frac{1}{3}-\frac{1}{2}-6\right)-0=-\frac{37}{6} \)

Let's go through each integral step by step, correcting any mistakes and ensuring clarity in the calculations. ### Problem 19 Evaluate the integral: \[ \int_{3}^{3}\left(x^{2}+2 x-4\right) d x \] 1. **Evaluate the definite integral**: Since the limits of integration are the same (from 3 to 3), the value of the integral is: \[ \int_{3}^{3} f(x) \, dx = 0 \] ### Problem 21 Evaluate the integral: \[ \int_{0}^{2}\left(\frac{4}{5} t^{3}-\frac{4}{4} t^{2}+\frac{2}{5} t\right) d t \] 1. **Find the antiderivative**: \[ F(t) = \left[\frac{1}{3} \cdot \frac{4}{5} t^{4} - \frac{1}{4} t^{3} + \frac{2}{5} \cdot \frac{1}{2} t^{2}\right] = \left[\frac{4}{15} t^{4} - \frac{1}{4} t^{3} + \frac{1}{5} t^{2}\right] \] 2. **Evaluate from 0 to 2**: \[ F(2) = \frac{4}{15} \cdot 16 - \frac{1}{4} \cdot 8 + \frac{1}{5} \cdot 4 = \frac{64}{15} - 2 + \frac{4}{5} \] Convert \(2\) to a fraction with a common denominator: \[ 2 = \frac{30}{15}, \quad \frac{4}{5} = \frac{12}{15} \] Thus, \[ F(2) = \frac{64}{15} - \frac{30}{15} + \frac{12}{15} = \frac{46}{15} \] And \(F(0) = 0\). 3. **Final result**: \[ \int_{0}^{2} \left(\frac{4}{5} t^{3}-t^{2}+\frac{2}{5} t\right) dt = \frac{46}{15} \] ### Problem 23 Evaluate the integral: \[ \int_{1}^{3} \sqrt{x} \, dx \] 1. **Find the antiderivative**: \[ F(x) = \left[\frac{2}{3} x^{3/2}\right] \] 2. **Evaluate from 1 to 3**: \[ F(3) = \frac{2}{3} \cdot 3^{3/2} = \frac{2}{3} \cdot 3\sqrt{3} = 2\sqrt{3} \] \[ F(1) = \frac{2}{3} \cdot 1^{3/2} = \frac{2}{3} \] 3. **Final result**: \[ \int_{1}^{3} \sqrt{x} \, dx = 2\sqrt{3} - \frac{2}{3} \] ### Problem 25 Evaluate the integral: \[ \int_{\pi / 6}^{\pi} \sin \theta \, d \theta \] 1. **Find the antiderivative**: \[ F(\theta) = -\cos \theta \] 2. **Evaluate from \(\frac{\pi}{6}\) to \(\pi\)**: \[ F(\pi) = -\cos(\pi) = 1 \] \[ F\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \] 3. **Final result**: \[ \int_{\pi / 6}^{\pi} \sin \theta \, d \theta = 1 - \left(-\frac{\sqrt{3}}{2}\right) = 1 + \frac{\sqrt{3}}{2} \] ### Problem 27 Evaluate the integral: \[ \int_{0}^{1}(u+2)(u-3) \, du \] 1. **Expand the integrand**: \[ (u+2)(u-3) = u^2 - 3u + 2u - 6 = u^2 - u - 6 \] 2. **Find the antiderivative**: \[ F(u) = \left[\frac{1}{3} u^{3} - \frac{1}{2} u^{2} - 6u\right] \] 3. **Evaluate from 0 to 1**: \[ F(1) = \frac{1}{3} - \frac{1}{2} - 6 = \frac{1}{3} - \frac{3}{6} - \frac{36}{6} = \frac{1}{3} - \frac{39}{6} \] Convert \(\frac{1}{3}\) to sixths: \[ \frac{1}{3} = \frac{2}{6} \] Thus, \[ F(1) = \frac{2}{6} - \frac{39}{6} = -\frac{37}{6} \] 4. **Final result**: \[ \int_{0}^{1}(u+2)(u-3) \, du = -\frac{37}{6} \] ### Summary of Results - Problem 19: \(0\) - Problem 21: \(\frac{46}{15}\) - Problem 23: \(2\sqrt{3} - \frac{2}{3}\) - Problem 25: \(1 + \frac{\sqrt{3}}{2}\) - Problem 27: \(-\frac{37}{6}\)