question 4\n(a) determine the following integrals:\n(i) $int\frac{4x^{3}-x}{4x^{4}}dx$\n(ii) $int xe^{(4…

question 4\n(a) determine the following integrals:\n(i) $int\frac{4x^{3}-x}{4x^{4}}dx$\n(ii) $int xe^{(4 - x^{2})}dx$\n(iii) $int(cos2x+sin x)dx$\n(b) determine the area enclosed by $y = x^{2}+2x - 3$ and the $x$-axis between $x = 0$ and $x = 2$.

question 4\n(a) determine the following integrals:\n(i) $int\frac{4x^{3}-x}{4x^{4}}dx$\n(ii) $int xe^{(4 - x^{2})}dx$\n(iii) $int(cos2x+sin x)dx$\n(b) determine the area enclosed by $y = x^{2}+2x - 3$ and the $x$-axis between $x = 0$ and $x = 2$.

Answer

Explanation:

Step1: Simplify the first integral

Rewrite $\int\frac{4x^{3}-x}{4x^{4}}dx$ as $\int(\frac{4x^{3}}{4x^{4}}-\frac{x}{4x^{4}})dx=\int(\frac{1}{x}-\frac{1}{4x^{3}})dx$.

Step2: Integrate term - by - term

$\int\frac{1}{x}dx-\int\frac{1}{4x^{3}}dx=\ln|x|+\frac{1}{8x^{2}}+C$.

Step3: Solve the second integral using substitution

Let $u = 4 - x^{2}$, then $du=-2xdx$ and $xdx=-\frac{1}{2}du$. So $\int xe^{(4 - x^{2})}dx=-\frac{1}{2}\int e^{u}du=-\frac{1}{2}e^{u}+C=-\frac{1}{2}e^{4 - x^{2}}+C$.

Step4: Integrate the third integral term - by - term

$\int(\cos2x+\sin x)dx=\int\cos2x dx+\int\sin x dx$. Since $\int\cos2x dx=\frac{1}{2}\sin2x$ and $\int\sin x dx=-\cos x$, the result is $\frac{1}{2}\sin2x-\cos x + C$.

Step5: Find the area for part (b)

First, find where $y = x^{2}+2x - 3=(x + 3)(x - 1)$ intersects the $x$ - axis. The roots are $x=-3$ and $x = 1$. For $x\in[0,2]$, we need to consider the sign of $y$. When $x\in[0,1]$, $y\leqslant0$ and when $x\in[1,2]$, $y\geqslant0$. The area $A=\left|\int_{0}^{1}(x^{2}+2x - 3)dx\right|+\int_{1}^{2}(x^{2}+2x - 3)dx$. $\int(x^{2}+2x - 3)dx=\frac{1}{3}x^{3}+x^{2}-3x+C$. $\int_{0}^{1}(x^{2}+2x - 3)dx=\left(\frac{1}{3}+1 - 3\right)-0=-\frac{5}{3}$, so $\left|\int_{0}^{1}(x^{2}+2x - 3)dx\right|=\frac{5}{3}$. $\int_{1}^{2}(x^{2}+2x - 3)dx=\left(\frac{8}{3}+4 - 6\right)-\left(\frac{1}{3}+1 - 3\right)=\frac{8 + 12-18}{3}-\frac{1 + 3 - 9}{3}=\frac{2}{3}+\frac{5}{3}=\frac{7}{3}$. $A=\frac{5}{3}+\frac{7}{3}=4$.

Answer:

(a)(i) $\ln|x|+\frac{1}{8x^{2}}+C$ (a)(ii) $-\frac{1}{2}e^{4 - x^{2}}+C$ (a)(iii) $\frac{1}{2}\sin2x-\cos x + C$ (b) $4$