QUESTION 2 Determine $$ \int y d x $$ if: $$ 2.1 \quad y=\frac{x^{2}}{e^{-3 x}} $$ $$ 2.2=\left(2 \sin \frac{3}{2} x \cos \frac{3}{2} x\right)^{4} $$ 2.3 2.4 2.5 2.5 $$ y=\ln \left(\frac{x}{3}\right) $$

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## 2.1  
We are given  
$$
y=\frac{x^2}{e^{-3x}},
$$  
which can be rewritten as  
$$
y=x^2e^{3x}.
$$  
We need to compute  
$$
\int x^2e^{3x}\,dx.
$$

**First integration by parts:**  
Let  
$$
u=x^2,\quad dv=e^{3x}\,dx.
$$  
Then,  
$$
du=2x\,dx,\quad v=\frac{1}{3}e^{3x}.
$$  
Thus,  
$$
\int x^2e^{3x}\,dx = x^2\cdot\frac{1}{3}e^{3x} - \int\frac{1}{3}e^{3x}(2x)\,dx = \frac{x^2e^{3x}}{3}-\frac{2}{3}\int xe^{3x}\,dx.
$$

**Second integration by parts:**  
Now, evaluate  
$$
\int xe^{3x}\,dx.
$$  
Let  
$$
u=x,\quad dv=e^{3x}\,dx.
$$  
Then,  
$$
du=dx,\quad v=\frac{1}{3}e^{3x}.
$$  
So,  
$$
\int xe^{3x}\,dx = x\cdot\frac{1}{3}e^{3x} - \int \frac{1}{3}e^{3x}\,dx 
=\frac{xe^{3x}}{3}-\frac{1}{3}\cdot\frac{1}{3}e^{3x}
=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}.
$$

Now, substitute back into the previous result:
$$
\int x^2e^{3x}\,dx = \frac{x^2e^{3x}}{3}-\frac{2}{3}\left(\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}\right).
$$
Simplify the expression:
$$
\int x^2e^{3x}\,dx = \frac{x^2e^{3x}}{3}-\frac{2xe^{3x}}{9}+\frac{2e^{3x}}{27}.
$$
Factor out $$e^{3x}$$:
$$
\int x^2e^{3x}\,dx = e^{3x}\left(\frac{x^2}{3}-\frac{2x}{9}+\frac{2}{27}\right)+C.
$$
Alternatively, writing with a common denominator:
$$
\int x^2e^{3x}\,dx = \frac{e^{3x}}{27}\Bigl(9x^2-6x+2\Bigr)+C.
$$

---

## 2.2  
We are given  
$$
y=\Bigl(2\sin\frac{3}{2}x\cos\frac{3}{2}x\Bigr)^4.
$$  
Notice that  
$$
2\sin\frac{3}{2}x\cos\frac{3}{2}x=\sin3x.
$$  
Thus,  
$$
y=(\sin3x)^4.
$$
We need to compute  
$$
\int (\sin3x)^4\,dx.
$$

Use the power-reduction formula for $$\sin^4 u$$:  
$$
\sin^4 u=\frac{3-4\cos2u+\cos4u}{8}.
$$
With $$u=3x$$, we have:
$$
(\sin3x)^4=\frac{3-4\cos6x+\cos12x}{8}.
$$
Then,  
$$
\int (\sin3x)^4\,dx=\frac{1}{8}\int\Bigl(3-4\cos6x+\cos12x\Bigr)\,dx.
$$

Integrate term by term:
$$
\int 3\,dx=3x,
$$
$$
\int \cos6x\,dx=\frac{\sin6x}{6},
$$
$$
\int \cos12x\,dx=\frac{\sin12x}{12}.
$$

Therefore,
$$
\int (\sin3x)^4\,dx=\frac{1}{8}\left(3x-4\cdot\frac{\sin6x}{6}+\frac{\sin12x}{12}\right)+C.
$$
Simplify the coefficients:
$$
4\cdot\frac{1}{6}=\frac{2}{3}.
$$
So the result is:
$$
\int (\sin3x)^4\,dx=\frac{1}{8}\left(3x-\frac{2\sin6x}{3}+\frac{\sin12x}{12}\right)+C.
$$
Or, distributing $$\frac{1}{8}$$:
$$
\int (\sin3x)^4\,dx=\frac{3x}{8}-\frac{\sin6x}{12}+\frac{\sin12x}{96}+C.
$$

---

## 2.5  
We are given  
$$
y=\ln\left(\frac{x}{3}\right).
$$
We need to evaluate  
$$
\int \ln\left(\frac{x}{3}\right)\,dx.
$$

Use integration by parts with:
$$
u=\ln\left(\frac{x}{3}\right),\quad dv=dx.
$$
Then,
$$
du=\frac{d}{dx}\ln\left(\frac{x}{3}\right)dx=\frac{1}{x}\,dx,\quad v=x.
$$
Thus,
$$
\int \ln\left(\frac{x}{3}\right)\,dx = x\ln\left(\frac{x}{3}\right)-\int x\cdot\frac{1}{x}\,dx = x\ln\left(\frac{x}{3}\right)-\int 1\,dx.
$$
Finally,
$$
\int \ln\left(\frac{x}{3}\right)\,dx = x\ln\left(\frac{x}{3}\right)-x+C.
$$
$$ \int y\,dx = \frac{e^{3x}}{27}(9x^2 - 6x + 2) + C $$ $$ \int (\sin3x)^4\,dx = \frac{3x}{8} - \frac{\sin6x}{12} + \frac{\sin12x}{96} + C $$ $$ \int \ln\left(\frac{x}{3}\right)\,dx = x\ln\left(\frac{x}{3}\right) - x + C $$

QUESTION 2
Determine if:

2.3
2.4
2.5
2.5

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QUESTION 2 Determine if: 2.3 2.4 2.5 2.5