6. $int_{0}^{1}\frac{x^{3}+1}{x + 1}dx$\n7. $int_{-2}^{2}|x - 3|dx$\n8. $int_{0}^{5}sqrt{2x+4}dx$\n9…

Explanation:

Step1: Simplify the integrand for problem 6

First, simplify $\frac{x^{3}+1}{x + 1}$. Since $x^{3}+1=(x + 1)(x^{2}-x + 1)$, then $\frac{x^{3}+1}{x + 1}=x^{2}-x + 1$. [ \begin{align*} \int_{0}^{1}(x^{2}-x + 1)dx&=\left[\frac{1}{3}x^{3}-\frac{1}{2}x^{2}+x\right]_{0}^{1}\ &=\left(\frac{1}{3}\times1^{3}-\frac{1}{2}\times1^{2}+1\right)-\left(\frac{1}{3}\times0^{3}-\frac{1}{2}\times0^{2}+0\right)\ &=\frac{1}{3}-\frac{1}{2}+1\ &=\frac{2 - 3+6}{6}=\frac{5}{6} \end{align*} ]

Step2: Split the integral for problem 7

For $y = |x - 3|$, when $x<3$, $|x - 3|=3 - x$; when $x\geq3$, $|x - 3|=x - 3$. [ \begin{align*} \int_{-2}^{2}|x - 3|dx&=\int_{-2}^{2}(3 - x)dx\ &=\left[3x-\frac{1}{2}x^{2}\right]_{-2}^{2}\ &=(3\times2-\frac{1}{2}\times2^{2})-(3\times(-2)-\frac{1}{2}\times(-2)^{2})\ &=(6 - 2)-(-6 - 2)\ &=4 + 8=12 \end{align*} ]

Step3: Use substitution for problem 8

Let $u = 2x+4$, then $du=2dx$. When $x = 0$, $u = 4$; when $x = 5$, $u=14$. [ \begin{align*} \int_{0}^{5}\sqrt{2x + 4}dx&=\frac{1}{2}\int_{4}^{14}u^{\frac{1}{2}}du\ &=\frac{1}{2}\times\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{4}^{14}\ &=\frac{1}{3}(14^{\frac{3}{2}}-4^{\frac{3}{2}})\ &=\frac{1}{3}(14\sqrt{14}-8) \end{align*} ]

Step4: Evaluate the sine - integral for problem 9

[ \begin{align*} \int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\sin xdx&=-\cos x\big|_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\ &=-\cos\frac{5\pi}{2}+\cos\frac{\pi}{2}\ &=0 \end{align*} ]

Step5: Use the double - angle formula for problem 10

Since $\cos^{2}x=\frac{1+\cos(2x)}{2}$, then $3\cos^{2}x=\frac{3}{2}(1 + \cos(2x))$. [ \begin{align*} \int_{0}^{\pi}3\cos^{2}xdx&=\int_{0}^{\pi}\frac{3}{2}(1+\cos(2x))dx\ &=\frac{3}{2}\left[x+\frac{1}{2}\sin(2x)\right]_{0}^{\pi}\ &=\frac{3}{2}(\pi+0)-(0 + 0)\ &=\frac{3\pi}{2} \end{align*} ]

Answer:

6. $\frac{5}{6}$
7. $12$
8. $\frac{1}{3}(14\sqrt{14}-8)$
9. $0$
10. $\frac{3\pi}{2}$