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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 2^2... 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

Substituting u=x3u = x - 3, the limits change from x=3x = 3 to u=0u = 0 and from x=4x = 4 to u=1u = 1.

Thus, we have: 01(u+5)u du\int_{0}^{1} (u + 5)\sqrt{u} \ du.

Now, we can rewrite this integral as: 01(u+5)u1/2 du=01(u3/2+5u1/2) du\int_{0}^{1} (u + 5)u^{1/2} \ du = \int_{0}^{1} (u^{3/2} + 5u^{1/2}) \ du.

Calculating each integral:

  1. For u3/2u^{3/2}: u3/2 du=u5/25/2=25u5/2\int u^{3/2} \ du = \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2} evaluated from 0 to 1 gives: 25.\frac{2}{5}.

  2. For 5u1/25u^{1/2}: 5u1/2 du=5u3/23/2=103u3/25 \int u^{1/2} \ du = 5 \cdot \frac{u^{3/2}}{3/2} = \frac{10}{3}u^{3/2} evaluated from 0 to 1 gives: 103.\frac{10}{3}.

Adding both results together: 25+103=6+5015=5615.\frac{2}{5} + \frac{10}{3} = \frac{6 + 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: For n=1n=1, the left-hand side (LHS) is: 1×22=2,1 \times 2^2 = 2,
while the right-hand side (RHS) is: 2+(11)21+1=2.2 + (1 - 1)2^{1 + 1} = 2.
Thus, LHS = RHS for n=1n=1.

Inductive Step: Assume true for n=kn=k: LHS = (1×22)+(2×22)++(k×2k)=2+(k1)2k+1(1 \times 2^2) + (2 \times 2^2) + \cdots + (k \times 2^k) = 2 + (k - 1)2^{k + 1}.

For n=k+1n=k+1, we have: LHS=(1×22)+(2×22)++(k×2k)+((k+1)×2k+1)\text{LHS} = (1 \times 2^2) + (2 \times 2^2) + \cdots + (k \times 2^k) + ((k + 1) \times 2^{k + 1}) Substituting the inductive assumption: =(2+(k1)2k+1)+((k+1)×2k+1)=2+(k+11)2k+2=2+k2k+2.= (2 + (k - 1)2^{k + 1}) + ((k + 1) \times 2^{k + 1}) = 2 + (k + 1 - 1)2^{k + 2} = 2 + k2^{k + 2}.
Thus, true for n=k+1n=k+1. Therefore, by induction, the statement is true for all integers 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

Let XX be the number of treadmills in use. Since each treadmill is used independently with a probability of 0.650.65, the expression becomes: P(X=3)=(53)(0.65)3(0.35)2.P(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

The expression changes to incorporate the fact that all rowing machines are not in use, which occurs with a probability of (0.4)0(0.4)^0. Thus the expression becomes: P(X=3,Y=0)=(53)(0.65)3(0.35)2×(0.4)0=(53)(0.65)3(0.35)2.P(X=3, Y=0) = {5 \choose 3} (0.65)^3 (0.35)^{2} \times (0.4)^0 = {5 \choose 3} (0.65)^3 (0.35)^{2}.

Step 5

Find ONE possible set of values for $p$ and $q$ such that $$2022C_{80} + 2022C_{81} + 2022C_{193} = PC_q.$$

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Answer

Using the identity, we can express: 2022C80+2022C81=2022C81+2022C80+2022C193.2022C_{80} + 2022C_{81} = 2022C_{81} + 2022C_{80} + 2022C_{193}.

By simplification, we find: 2022C81+2022C80+2022C193=2022C80.2022C_{81} + 2022C_{80} + 2022C_{193} = 2022C_{80}.
Thus, taking p=2024p=2024 and q=81q=81 gives a valid solution.

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