Let $f(x) = 3x^2 + x$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2001 - Paper 1

To evaluate this integral, use the substitution $u = 1 + e^x$, hence $du = e^x dx$.

Then the integral becomes:

$\int \frac{1}{u} du = \ln |u| + C = \ln |1 + e^x| + C.$

Evaluate this integral as follows:

$\int 3 \cos x dx = 3 \sin x + C.$

Apply the limits:

$\left[3 \sin x\right]_0^2 = 3 \sin(2) - 3 \sin(0) = 3 \sin(2).$

The word 'ALGE BRAIC' has 9 letters with A occurring twice and I occurring twice.

The formula for arrangements is:

$\frac{n!}{n_1! \cdot n_2!},$ where $n$ is the total number of letters, $n_1$ and $n_2$ are the counts of respective repeated letters.

Thus:

$\frac{9!}{2! \cdot 2!} = \frac{362880}{4} = 90720.$

The vowels A, A, E, I need to occupy positions 2, 3, 5, and 8. The arrangements of the vowels:

$\frac{4!}{2!} = 12.$

The remaining consonants L, G, B, R, C can occupy the remaining 5 positions:

$5! = 120.$

Therefore, the total arrangements is:

$12 \times 120 = 1440.$

Using the binomial theorem, the general term is given by:

$T_k = \binom{n}{k} (x^2)^{n-k} (-1)^k = \binom{9}{k} x^{2(n-k)}(-1)^k.$

We seek the term where the exponent of $x$ is zero:

$2(n-k) = 0 \Rightarrow n-k = 0 \Rightarrow k = 9.$

Thus, we find the term:

inom{9}{9}(-1)^9 = -1.