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Evaluate $$\int_{3}^{4} (x + 2) \sqrt{x - 3} \, dx$$ using the substitution $u = x - 3$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2023 - Paper 1

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Evaluate-$$\int_{3}^{4}-(x-+-2)-\sqrt{x---3}-\,-dx$$-using-the-substitution-$u-=-x---3$-HSC-SSCE Mathematics Extension 1-Question 12-2023-Paper 1.png

Evaluate $$\int_{3}^{4} (x + 2) \sqrt{x - 3} \, dx$$ using the substitution $u = x - 3$. (b) Use mathematical induction to prove that $$(1 \times 2^2) + (2 \times 2... show full transcript

Worked Solution & Example Answer:Evaluate $$\int_{3}^{4} (x + 2) \sqrt{x - 3} \, dx$$ using the substitution $u = x - 3$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2023 - Paper 1

Step 1

Evaluate $$\int_{3}^{4} (x + 2) \sqrt{x - 3} \, dx$$ using the substitution $u = x - 3$.

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Answer

To evaluate the integral, first we substitute:

u=x3dx=du;When x=3,  u=0 and when x=4,  u=1.u = x - 3 \Rightarrow dx = du;\quad \text{When } x = 3, \; u = 0 \text{ and when } x = 4, \; u = 1.

Thus, the integral becomes:

01(u+5)udu.\int_0^1 (u + 5) \sqrt{u} \, du.

Now we calculate:

=01(u+5)u1/2du=01(u3/2+5u1/2)du = \int_0^1 (u + 5) \cdot u^{1/2} \, du = \int_0^1 (u^{3/2} + 5u^{1/2}) \, du

Evaluating the integrals:

[u5/25/2+5u3/23/2]01\left[ \frac{u^{5/2}}{5/2} + 5 \cdot \frac{u^{3/2}}{3/2} \right]_0^1

=[2515/2+10313/2]0=25+103=6+5015=5615.= \left[ \frac{2}{5} \cdot 1^{5/2} + \frac{10}{3} \cdot 1^{3/2} \right] - 0 = \frac{2}{5} + \frac{10}{3} = \frac{6 + 50}{15} = \frac{56}{15}.

Thus, the value of the integral is: 5615\frac{56}{15}.

Step 2

Use mathematical induction to prove that $(1 \times 2^2) + (2 \times 2^2) + \cdots + (n \times 2^n) = 2 + (n - 1)2^{n + 1}$ for all integers $n \geq 1$.

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Answer

Base Case:

Let n=1n = 1:

LHS=1×22=2;RHS=2+(11)21+1=2.LHS = 1 \times 2^2 = 2; \quad RHS = 2 + (1 - 1)2^{1 + 1} = 2. Thus, LHS=RHSLHS = RHS holds for n=1n = 1.

Inductively Assume:

Assume it holds for n=kn = k:

(1×22)++(k×2k)=2+(k1)2k+1.(1 \times 2^2) + \ldots + (k \times 2^k) = 2 + (k - 1)2^{k + 1}.

Inductive Step:

For n=k+1n = k + 1:

LHS=(1×22)++(k×2k)+((k+1)×2k+1).LHS = (1 \times 2^2) + \ldots + (k \times 2^k) + ((k+1) \times 2^{k+1}). Using the inductive hypothesis: =2+(k1)2k+1+(k+1)2k+1=2+k2k+1.= 2 + (k - 1)2^{k + 1} + (k + 1)2^{k + 1} = 2 + k 2^{k + 1} .

=2+(k+11)2(k+1+1).= 2 + (k + 1 - 1)2^{(k + 1 + 1)}. Thus, LHS=RHS, so the statement holds for n=k+1.LHS = RHS, \text{ so the statement holds for } n = k + 1.

By induction, it holds for all n1n \geq 1.

Step 3

Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use.

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Answer

The probability of exactly 3 out of 5 treadmills being used can be calculated using the binomial distribution:

P(X=k)=(nk)pk(1p)nk,P(X = k) = {n \choose k} p^k (1 - p)^{n - k}, where:

  • n=5n = 5 (total treadmills),
  • k=3k = 3 (treadmills in use),
  • p=0.65p = 0.65 (probability of each treadmill being used).

Thus, P(X=3)=(53)(0.65)3(0.35)53P(X = 3) = {5 \choose 3} (0.65)^3 (0.35)^{5-3}

Calculating:

=10(0.65)3(0.35)2. = 10 \cdot (0.65)^3 \cdot (0.35)^2.

Step 4

Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use and no rowing machines are in use.

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Answer

Likewise, the probability of exactly 3 treadmills in use:

Using results from (c)(i): P(T=3)=(53)(0.65)3(0.35)2.P(T = 3) = {5 \choose 3} (0.65)^3 (0.35)^{2}.

For the 4 rowing machines being unused: P(R=0)=(0.4)0(0.6)4.P(R = 0) = (0.4)^{0}(0.6)^{4}.

Thus, the overall probability: P=P(T=3)P(R=0)=(53)(0.65)3(0.35)2(0.6)4.P = P(T=3) \cdot P(R=0) = {5 \choose 3} (0.65)^3 (0.35)^{2} \cdot (0.6)^4.

Step 5

Find ONE possible set of values for $p$ and $q$ such that $2022C_{80} + 2022C_{81} + 2022C_{93} = PC_{q}$.

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Answer

By using the identity,

2022C80+2022C81=2022C81+1.2022C_{80} + 2022C_{81} = 2022C_{81 + 1}.

Also,

2022C81+1+2022C93=2022C80+93.2022C_{81 + 1} + 2022C_{93} = 2022C_{80 + 93}.

This leads to: 2022C80+2022C81+2022C93=2022C81.2022C_{80} + 2022C_{81} + 2022C_{93} = 2022C_{81}.

Thus, a possible solution is (p,q)=(2024,81).(p, q) = (2024, 81).

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