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

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

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

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

Step 1

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

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Answer

To solve the integral, we will use the substitution ( u = x - 3 ). Thus, the bounds change as follows: when ( x = 3 ), ( u = 0 ) and when ( x = 4 ), ( u = 1 ).

The integral becomes:

01(u+5)3(u+3)du=01(u+5)0udu\int_{0}^{1} (u + 5)\sqrt{3 - (u + 3)} \, du = \int_{0}^{1} (u + 5)\sqrt{0 - u} \, du

Next, we'll simplify and compute this integral. We have:

01(u+5)(u)du=01(uu+5u)du\int_{0}^{1} (u + 5)(\sqrt{-u}) \, du = \int_{0}^{1} (-u\sqrt{u} + 5\sqrt{u}) \, du

Breaking that down:

  1. Calculate ( \int_{0}^{1} -u\sqrt{u} , du ):

    • This can be computed as ( \int_{0}^{1} -u^{3/2} , du = -\frac{2}{5}u^{5/2} \Big|_{0}^{1} = -\frac{2}{5} )
  2. Calculate ( \int_{0}^{1} 5\sqrt{u} , du ):

    • This becomes ( 5\cdot \frac{2}{3}u^{3/2} \Big|_{0}^{1} = \frac{10}{3} )

Combining both results, we get:

=25+103=541015=4415.= -\frac{2}{5} + \frac{10}{3} = \frac{54-10}{15} = \frac{44}{15}.

Step 2

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

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Answer

Base Case (n=1): Evaluating the left-hand side: ( LHS = 1 \times 2^2 = 2 ) For the right-hand side: ( RHS = 2 + (1-1)2^{1+1} = 2 ) Both sides are equal.

Inductive Step: Assume true for ( n = k ): ( (1 \times 2^2) + (2 \times 2^2) + \cdots + (k \times 2^k) = 2 + (k-1)2^{k+1} ) Now prove for ( n = k+1 ): ( LHS = (1 \times 2^2) + (2 \times 2^2) + \cdots + (k \times 2^k) + ((k+1) \times 2^{k+1}) ) ( = (2 + (k-1)2^{k+1}) + (k+1)(2^{k+1}) ) ( = 2 + k \times 2^{k+1} + (k+1) \times 2^{k+1} = 2 + (k+1 - 1) \times 2^{k+2} ) ( = 2 + k 2^{k+2} ) Thus, 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

Given that each treadmill is used 65% of the time, we can model this situation using the binomial probability formula:

P(X=k)=C(n,k)pk(1p)(nk)P(X = k) = C(n, k) p^k (1-p)^{(n-k)}

Here, ( n = 5 ) (the number of treadmills), ( k = 3 ), and ( p = 0.65 ). Thus, the probability is given by:

P(X=3)=C53(0.65)3(0.35)2P(X = 3) = C_{5}^{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

For this scenario, we will find the probability for exactly 3 treadmills and 0 rowing machines. The treadmills probability is already calculated as:

C53(0.65)3(0.35)2C_{5}^{3} (0.65)^{3} (0.35)^{2}

For the rowing machines, the probability of 0 machines being used (out of 4 available, with use probability of 0.40) is:

Combining both expressions:

P=C53(0.65)3(0.35)2×(0.60)4P = C_{5}^{3} (0.65)^{3} (0.35)^{2} \times (0.60)^{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

From the given equation, we can use the property of binomial coefficients:

Crn+Crn+1=Cr+1n+1C_{r}^{n} + C_{r}^{n+1} = C_{r+1}^{n+1}

Therefore:

2022C80+2022C81=2022C82 (combining the first two terms)2022C_{80} + 2022C_{81} = 2022C_{82} \text{ (combining the first two terms)}

Now we can simplify it as follows:

2022C82+2022C93=2022C80+2022C81+2022C93=PCq2022C_{82} + 2022C_{93} = 2022C_{80} + 2022C_{81} + 2022C_{93} = PC_{q}

Comparing terms yields: ( p = 2022, q = 81 ) is a possible solution.

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