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Find \( \int x \ln x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2011 - Paper 1
Question 1
Find \( \int x \ln x \, dx \). Evaluate \( \int_{0}^{3} \sqrt{x + 1} \, dx \). (c) Find real numbers \( a, b \) and \( c \) such that \[ \frac{1}{x^{2}(x - 1)} = \... show full transcript
Worked Solution & Example Answer:Find \( \int x \ln x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2011 - Paper 1
Step 1
Find \( \int x \ln x \, dx \)
Answer
To solve this integral, we will use integration by parts, where we let:
- ( u = \ln x )
- ( dv = x , dx )
Then we find ( du = \frac{1}{x} , dx ) and ( v = \frac{x^{2}}{2} ). Using the integration by parts formula, ( \int u , dv = uv - \int v , du ), we have:
[\int x \ln x , dx = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} \cdot \frac{1}{x} , dx]
This simplifies to:
[\frac{x^{2}}{2} \ln x - \frac{1}{2} \int x , dx]
Calculating ( \int x , dx ) gives ( \frac{x^{2}}{2} ), leading to:
[\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + C]
Step 2
Evaluate \( \int_{0}^{3} \sqrt{x + 1} \, dx \)
Answer
We can use the substitution ( u = x + 1 ), hence ( du = dx ) and change the limits accordingly:
When ( x = 0, u = 1 ) and when ( x = 3, u = 4 ).
Thus, the integral becomes:
[\int_{1}^{4} \sqrt{u} , du = \frac{2}{3} u^{\frac{3}{2}} \bigg|_{1}^{4} = \frac{2}{3} (8 - 1) = \frac{14}{3}]
Step 3
Find real numbers \( a, b \) and \( c \) such that \( \frac{1}{x^{2}(x - 1)} = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x - 1} \)
Answer
To find the values of ( a, b, ) and ( c ), we multiply both sides by ( x^{2}(x - 1) ):
[ 1 = a x (x - 1) + b (x - 1) + c x^{2} ]
Expanding and equating coefficients gives us a system of equations:
- ( a + c = 0 ) (for ( x^{2} ))
- ( -a + b = 0 ) (for ( x ))
- ( -b = 1 ) (constant term)
Solving these, we find:
- ( c = -1 )
- ( b = -1 )
- ( a = -1 )
Step 4
Step 5
Find \( \int \cos^{3} \theta \, d\theta \)
Answer
To solve this integral, we can use the identity:
[ \cos^{3} \theta = \cos \theta (1 - \sin^{2} \theta) ]
Thus, the integral becomes:
[ \int \cos^{3} \theta , d\theta = \int \cos \theta , d\theta - \int \cos \theta , \sin^{2} \theta , d\theta ]
Using substitution, ( u = \sin \theta ) gives: [ du = \cos \theta , d\theta], hence:
[ \int (1 - u^{2}) , du = u - \frac{u^{3}}{3} + C = \sin \theta - \frac{\sin^{3} \theta}{3} + C ]
Step 6
Evaluate \( \int_{-1}^{1} \frac{1}{5 - 2t^{2}} \, dt \)
Answer
This integral can be evaluated using symmetry. Since the integrand is an even function, we have:
[ \int_{-1}^{1} f(t) , dt = 2 \int_{0}^{1} f(t) , dt ]
Calculating the integral: [= 2 \int_{0}^{1} \frac{1}{5 - 2t^{2}} , dt ].
Using the substitution ( u = 5 - 2t^{2} ), we find the definite integral results to: [= \frac{\pi}{\sqrt{10}}]