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Use integration by parts to find \( \int xe^x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2004 - Paper 1
Question 1
Use integration by parts to find \( \int xe^x \, dx \). (b) Evaluate \( \int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos^3 x} \, dx \). (c) By completing the square, find... show full transcript
Worked Solution & Example Answer:Use integration by parts to find \( \int xe^x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2004 - Paper 1
Step 1
Use integration by parts to find \( \int xe^x \, dx \).
Answer
To solve ( \int xe^x , dx ), we apply integration by parts, using the formula:
[ \int u , dv = uv - \int v , du. ]
Letting ( u = x ) and ( dv = e^x , dx ), we find:
- ( du = dx )
- ( v = e^x )
Then, substituting into the formula gives:
[ \int xe^x , dx = x e^x - \int e^x , dx = x e^x - e^x + C. ]
Thus, the final answer is:
[ \int xe^x , dx = (x - 1)e^x + C. ]
Step 2
Evaluate \( \int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos^3 x} \, dx \).
Answer
To evaluate this integral, we can use the substitution ( u = \cos x ), which gives ( du = -\sin x , dx ). Changing the limits accordingly:
When ( x = 0 ), ( u = 1 ) and when ( x = \frac{\pi}{2} ), ( u = 0 ).
Thus, the integral becomes:
[ -\int_1^0 \frac{1}{u^3} (-du) = \int_0^1 \frac{1}{u^3} , du. ]
Computing the integral:
[ \int_0^1 u^{-3} , du = \left[ \frac{u^{-2}}{-2} \right]_0^1 = 0 - (-\frac{1}{2}) = \frac{1}{2}. ]
Therefore, the answer is ( \frac{1}{2} ).
Step 3
By completing the square, find \( \int \frac{dx}{\sqrt{5 + 4x - x^2}} \).
Answer
To complete the square for the expression ( 5 + 4x - x^2 ), we rewrite it as:
[ - (x^2 - 4x - 5) = -\left((x - 2)^2 - 4 - 5\right) = -\left((x - 2)^2 - 9\right) = 9 - (x - 2)^2. ]
Therefore, the integral becomes:
[ \int \frac{dx}{\sqrt{9 - (x - 2)^2}}. ]
This presents a standard form for arcsine:
[ = 2 \arcsin\left(\frac{x - 2}{3}\right) + C. ]
Step 4
(i) Find real numbers \( a \) and \( b \) such that \( \frac{x^2 - 7x + 4}{(x + 1)(x - 1)^2} = \frac{a}{x+1} + \frac{b}{x - 1} - \frac{1}{(x - 1)^2}. \
Answer
We start with the equation:
[ x^2 - 7x + 4 = a(x-1)^2 + b(x + 1)(x - 1). ]
Expanding this gives:
[ ax^2 - 2ax + a + bx^2 - b. ]
Combining like terms, we must match coefficients:
[ (a + b)x^2 + (-2a + b)x + (a - b) = 1x^2 - 7x + 4. ]
From this, we have a system of equations:
- ( a + b = 1 )
- ( -2a + b = -7 )
- ( a - b = 4 )
Solving this system gives: ( a = 3 ) and ( b = -2 ).
Step 5
(ii) Hence find \( \int \frac{x^2 - 7x + 4}{(x + 1)(x - 1)^2} \, dx \).
Answer
Using the values found previously, we substitute:
[ \frac{3}{x+1} - \frac{2}{x-1} - \frac{1}{(x-1)^2}. ]
This results in three separate integrals:
[ 3 \int \frac{dx}{x+1} - 2 \int \frac{dx}{x-1} - \int \frac{dx}{(x-1)^2}. ]
Calculating gives:
[ 3 \ln|x+1| - 2 \ln|x-1| + \frac{1}{x-1} + C. ]
Step 6
Use the substitution \( x = 2 \sin \theta \) to find \( \int_0^{\frac{x^2}{4 - x^2}} \, dx \).
Answer
With the substitution ( x = 2 \sin \theta ), we have: [ dx = 2 \cos \theta , d\theta. ]
The integral limits change as follows: When ( x = 0 \rightarrow \theta = 0 ), and when ( x = 2 \rightarrow \theta = \frac{\pi}{2} ).
Thus, we rewrite the integral and simplify: [ 2 \int_0^\frac{\pi}{2} (2 \sin \theta) (2 \cos \theta) , d\theta. ] This simplifies to: [ 4 \int_0^\frac{\pi}{2} \sin \theta \cos \theta , d\theta = 4 \left[ \frac{1}{2} \sin^2 \theta \right]_0^\frac{\pi}{2} = 4 \cdot \frac{1}{2} = 2. ]
Therefore, the final result is ( 2 ).