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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... 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, we first make the substitution:

Let ( u = x - 3 ), then ( du = dx ) and when ( x = 3, u = 0 ); when ( x = 4, u = 1 ). Thus, we can rewrite the integral as:

01(u+5)udu\int_{0}^{1} (u + 5)\sqrt{u} \, du.

Next, we can express ( \sqrt{u} ) as ( u^{1/2} ): =01(u+5)u1/2du= \int_{0}^{1} (u + 5)u^{1/2} \, du =01(u3/2+5u1/2)du = \int_{0}^{1} (u^{3/2} + 5u^{1/2}) \, du.

Now, we can integrate each term:

  1. For ( u^{3/2} ): ( \int u^{3/2} , du = \frac{2}{5}u^{5/2} ) evaluated from 0 to 1 gives ( \frac{2}{5} ).
  2. For ( 5u^{1/2} ): ( 5\int u^{1/2} , du = 5 \cdot \frac{2}{3}u^{3/2} ) evaluated from 0 to 1 gives ( \frac{10}{3} ).

Thus, the integral evaluates to: =25+103=615+5015=5615= \frac{2}{5} + \frac{10}{3} = \frac{6}{15} + \frac{50}{15} = \frac{56}{15}.

Step 2

Use mathematical induction to prove that $$(1 \times 2^2) + (2 \times 2^2) + (3 \times 2^2) + \cdots + (n \times 2^n) = 2 + (n - 1)2^{n + 1}$$ for all integers $n \geq 1$

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Answer

Base Case:

When ( n = 1 ): LHS = ( 1 \times 2^2 = 2 ) and RHS = ( 2 + (1 - 1)2^{1 + 1} = 2 ).

Thus, the base case holds.

Inductive Step:

Assume true for ( n = k ): (1×22)+(2×22)++(k×22)=2+(k1)2k+1(1 \times 2^2) + (2 \times 2^2) + \dots + (k \times 2^2) = 2 + (k - 1)2^{k + 1}.

Then, for ( n = k + 1 ): LHS = ( 2 + (k \times 2^2) + (k + 1)2^{k + 1} = 2 + (k - 1)2^{k + 1} + (k + 1)2^{k + 1} = 2 + k2^{k + 1} + 2^{k + 1} = 2 + (k + 1 - 1)2^{(k + 1) + 1} )

Therefore, by mathematical induction, it is true for all integers ( n \geq 1 ).

Step 3

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

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Answer

We can apply the binomial probability formula:

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

Where ( n = 5 ) (the number of treadmills), ( k = 3 ), and ( p = 0.65 ):

So, the expression becomes: P(X=3)=(53)(0.65)3(0.35)2P(X = 3) = {5 \choose 3} (0.65)^3 (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

Using the same formula for the treadmills:

P(X=3)=(53)(0.65)3(0.35)2P(X = 3) = {5 \choose 3} (0.65)^3 (0.35)^2

For the rowing machines, since they must all be off, the probability is:

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