question 4\n4.1 determine the following integrals:\n4.1.1 $\\int(2x - 3)^4dx$ (3)\n4.1.2 $\\int\\frac{2x^{2}+…

Explanation:

Step1: Solve 4.1.1

Let $u = 2x - 3$, then $du=2dx$ and $dx=\frac{1}{2}du$. $$\int(2x - 3)^4dx=\frac{1}{2}\int u^4du$$ Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{1}{2}\times\frac{u^{5}}{5}+C=\frac{(2x - 3)^5}{10}+C$

Step2: Solve 4.1.2

Rewrite $\int\frac{2x^{2}+x + 1}{x^{2}}dx=\int(2+\frac{1}{x}+\frac{1}{x^{2}})dx$ Integrating term - by - term: $\int 2dx+\int\frac{1}{x}dx+\int x^{-2}dx=2x+\ln|x|-\frac{1}{x}+C$

Step3: Solve 4.1.3

Let $u=\cos x$, then $du=-\sin xdx$ and $\int\frac{\sin x}{\cos^{5}x}dx=-\int u^{-5}du$ Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n + 1}+C(n\neq-1)$, we get $-\frac{u^{-4}}{-4}+C=\frac{1}{4\cos^{4}x}+C$

Step4: Solve 4.1.4

Since $\sec^{2}x-1=\tan^{2}x$, then $\int\sqrt{\sec^{2}x - 1}dx=\int\tan xdx=-\ln|\cos x|+C$

Step5: Solve 4.2

First, rewrite the parabola $x = 8 + 2y-y^{2}=-(y^{2}-2y - 8)=-(y - 4)(y+2)$ The area $A=\int_{-1}^{3}(8 + 2y-y^{2})dy$ Using the power - rule $\int x^n dx=\frac{x^{n+1}}{n + 1}+C(n\neq - 1)$: [ \begin{align*} \int_{-1}^{3}(8 + 2y-y^{2})dy&=\left[8y+y^{2}-\frac{y^{3}}{3}\right]_{-1}^{3}\ &=(8\times3+3^{2}-\frac{3^{3}}{3})-(8\times(-1)+(-1)^{2}-\frac{(-1)^{3}}{3})\ &=(24 + 9-9)-(-8 + 1+\frac{1}{3})\ &=24-(-\frac{24 - 3+1}{3})\ &=24+\frac{20}{3}\ &=\frac{72 + 20}{3}=\frac{92}{3} \end{align*} ]

Answer:

4.1.1: $\frac{(2x - 3)^5}{10}+C$ 4.1.2: $2x+\ln|x|-\frac{1}{x}+C$ 4.1.3: $\frac{1}{4\cos^{4}x}+C$ 4.1.4: $-\ln|\cos x|+C$ 4.2: $\frac{92}{3}$