question 3\n3.1 determine the following integrals:\n3.1.1 $int (x^{3}-2)sqrt{x^{4}-8x} dx$\n3.1.2 $int…

Explanation:

3.1.1

Step1: Use substitution

Let $u = x^{4}-8x$, then $du=(4x^{3}-8)dx = 4(x^{3}-2)dx$, and $(x^{3}-2)dx=\frac{1}{4}du$. So, $\int(x^{3}-2)\sqrt{x^{4}-8x}dx=\frac{1}{4}\int\sqrt{u}du$.

Step2: Integrate $\sqrt{u}$

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ for $n\neq - 1$. Here $n=\frac{1}{2}$, so $\frac{1}{4}\int\sqrt{u}du=\frac{1}{4}\times\frac{2}{3}u^{\frac{3}{2}}+C=\frac{1}{6}(x^{4}-8x)^{\frac{3}{2}}+C$.

3.1.2

Step1: Use substitution

Let $t=\tan(2x)$, then $dt = 2\sec^{2}(2x)dx$, and $\sec^{2}(2x)dx=\frac{1}{2}dt$. So, $\int e^{\tan(2x)}\sec^{2}(2x)dx=\frac{1}{2}\int e^{t}dt$.

Step2: Integrate $e^{t}$

Since $\int e^{t}dt=e^{t}+C$, then $\frac{1}{2}\int e^{t}dt=\frac{1}{2}e^{\tan(2x)}+C$.

3.1.3

Step1: Simplify the integrand

Since $\ln(e^{a})=a$ for any real - number $a$, then $\ln(e^{2x - 1})=2x-1$. So, $\int\ln(e^{2x - 1})dx=\int(2x - 1)dx$.

Step2: Integrate term - by - term

$\int(2x - 1)dx=\int 2xdx-\int 1dx$. Using $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n = 1$ for the first integral and $n = 0$ for the second integral), we get $2\times\frac{x^{2}}{2}-x+C=x^{2}-x+C$.

3.1.4

Step1: Factor the denominator

$x^{2}-6x + 9=(x - 3)^{2}$, so the integrand is $\frac{6x-10}{(x - 3)^{2}(x + 1)}$. We write $\frac{6x-10}{(x - 3)^{2}(x + 1)}=\frac{A}{x - 3}+\frac{B}{(x - 3)^{2}}+\frac{C}{x + 1}$. $6x-10=A(x - 3)(x + 1)+B(x + 1)+C(x - 3)^{2}$. Let $x=3$, then $6\times3-10=B(3 + 1)$, $8 = 4B$, $B = 2$. Let $x=-1$, then $6\times(-1)-10=C(-1 - 3)^{2}$, $-16 = 16C$, $C=-1$. Expand the right - hand side: $A(x - 3)(x + 1)+B(x + 1)+C(x - 3)^{2}=A(x^{2}-2x-3)+Bx + B+C(x^{2}-6x + 9)=(A + C)x^{2}+(-2A + B-6C)x+(-3A + B + 9C)$. Comparing the coefficients of $x^{2}$: $A + C = 0$, since $C=-1$, then $A = 1$. So, $\frac{6x-10}{(x - 3)^{2}(x + 1)}=\frac{1}{x - 3}+\frac{2}{(x - 3)^{2}}-\frac{1}{x + 1}$. $\int\frac{6x-10}{(x - 3)^{2}(x + 1)}dx=\int\frac{1}{x - 3}dx+2\int\frac{1}{(x - 3)^{2}}dx-\int\frac{1}{x + 1}dx$. $\int\frac{1}{x - 3}dx=\ln|x - 3|$, $2\int\frac{1}{(x - 3)^{2}}dx=2\times\frac{(x - 3)^{-1}}{-1}=-\frac{2}{x - 3}$, $\int\frac{1}{x + 1}dx=\ln|x + 1|$. The integral is $\ln|x - 3|-\frac{2}{x - 3}-\ln|x + 1|+C=\ln\left|\frac{x - 3}{x + 1}\right|-\frac{2}{x - 3}+C$.

3.2

Step1: Set up the integral for work

Given $F=-kx$ with $k = 30$ N/m and $x$ is the displacement. The work $W=\int_{0}^{0.4}-kx dx$ (since the spring is stretched from $x = 0$ to $x=0.4$ m).

Step2: Integrate $-kx$

$\int_{0}^{0.4}-kx dx=-k\int_{0}^{0.4}x dx$. Using $\int xdx=\frac{x^{2}}{2}+C$, we have $-k\left[\frac{x^{2}}{2}\right]_{0}^{0.4}$. Substitute $k = 30$: $-30\times\frac{0.4^{2}}{2}-(-30\times\frac{0^{2}}{2})=-30\times\frac{0.16}{2}=-2.4$ J. The negative sign just indicates the direction of the force relative to the displacement. The magnitude of the work done is $2.4$ J.

Answer:

3.1.1: $\frac{1}{6}(x^{4}-8x)^{\frac{3}{2}}+C$ 3.1.2: $\frac{1}{2}e^{\tan(2x)}+C$ 3.1.3: $x^{2}-x+C$ 3.1.4: $\ln\left|\frac{x - 3}{x + 1}\right|-\frac{2}{x - 3}+C$ 3.2: $2.4$ J