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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
Question 12
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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 \( u = x - 3 \)
Answer
To solve the integral, we start by making the substitution:
Let ( u = x - 3 ). Therefore, ( x = u + 3 ) and ( dx = du ).
Next, we need to change the limits of integration:
- When ( x = 3 ), ( u = 0 ).
- When ( x = 4 ), ( u = 1 ).
Now we rewrite the integral:
[ \int_{0}^{1} ((u + 3) + 2)\sqrt{u} du = \int_{0}^{1} (u + 5)\sqrt{u} du ]
Next, we simplify and calculate the new integral:
[ \int_{0}^{1} (u + 5)u^{1/2} du = \int_{0}^{1} (u^{3/2} + 5u^{1/2}) du ]
Calculating the integral gives:
[ \frac{2}{5}u^{5/2} \bigg|{0}^{1} + 5 \cdot \frac{2}{3}u^{3/2} \bigg|{0}^{1} = \frac{2}{5} + \frac{10}{3} = \frac{2}{5} + \frac{50}{15} = \frac{16}{15} ]
Step 2
Use mathematical induction to prove that \[ (1 \times 2^{2}) + (2 \times 2^{2}) + (3 \times 2^{2}) + \cdots + (n \times 2^{n}) = 2 + (n-1)2^{n+1} \]
Answer
To prove by induction:
-
Base Case: For ( n = 1 ):
- LHS: ( 1 \times 2^{2} = 2 )
- RHS: ( 2 + (1-1)2^{2} = 2 )
- LHS = RHS, so the base case holds.
-
Inductive Step: Assume true for ( n = k ):
- LHS: ( (1 \times 2^{2}) + (2 \times 2^{2}) + \cdots + (k \times 2^{k}) = 2 + (k - 1)2^{k+1} )
- For ( n = k + 1 ):
- RHS: ( 2 + (k)2^{k+1} + (k + 1)2^{k + 1} = 2 + (k + 1 - 1)2^{k + 1} = 2 + k2^{k + 1} + (k + 1)2^{k + 1} )
- Therefore, inductively true for all integers ( n \geq 1 ).
Step 3
Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use.
Answer
To calculate this, we use the binomial probability formula:
[ P(X = k) = C(n, k) p^k (1 - p)^{n - k} ]
where ( n = 5 ), ( k = 3 ), ( p = 0.65 ).
Thus, we have:
[ P(X = 3) = C_{5}^{3} (0.65)^{3} (0.35)^{2} ]
where ( C_{5}^{3} = 10 ), leading to:
[ P(X = 3) = 10(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.
Answer
We can use the combined binomial probability for two independent events: [ P(X = 3 ext{ treadmills}, Y = 0 ext{ rowing machines}) = P(X = 3) imes P(Y = 0) ]
Using the previously calculated ( P(X = 3) ), and for ( Y = 0 ):
- Here, ( n = 4 ), ( k = 0 ), ( p = 0.4 )
Thus: [ P(Y = 0) = C_{4}^{0}(0.4)^{0}(0.6)^{4} = (0.6)^{4} ]
Therefore: [ P(X = 3, Y = 0) = 10(0.65)^{3}(0.35)^{2} imes (0.6)^{4} ]
Step 5
Find ONE possible set of values for \( p \) and \( q \) such that \[ 2022C_{80} + 2022C_{81} + 2022C_{193} = PC_{q} \]
Answer
Using the identity: [ C_{r}^{n} = C_{n-1}^{r-1} + C_{n-1}^{r} ] This provides the relation: [ 2022C_{80 + 81 + 193} = 2022C_{81} + 2022C_{193} ]
So we can set: ( p = 2022 ) and ( q = 193 ) as a possible solution.