If \( \int_a^x f(t)dt = g(x) \), which of the following is a primitive of \( f(x)g(x) \)?
A - HSC - SSCE Mathematics Extension 2 - Question 5 - 2022 - Paper 1
Question 5
If \( \int_a^x f(t)dt = g(x) \), which of the following is a primitive of \( f(x)g(x) \)?
A. \( \frac{1}{2}[f(x)]^2 \)
B. \( \frac{1}{2}[f'(x)]^2 \)
C. \( \frac{... show full transcript
Worked Solution & Example Answer:If \( \int_a^x f(t)dt = g(x) \), which of the following is a primitive of \( f(x)g(x) \)?
A - HSC - SSCE Mathematics Extension 2 - Question 5 - 2022 - Paper 1
Step 1
Identify the Function Relationship
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Answer
Given that ( \int_a^x f(t)dt = g(x) ), we can apply the Fundamental Theorem of Calculus. It states that if ( G(x) = \int_a^x f(t)dt ), then ( G'(x) = f(x) ). Therefore, ( g'(x) = f(x) ).
Step 2
Finding the Primitive of \( f(x)g(x) \)
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Answer
We want to find a primitive of ( f(x)g(x) ). Using the product rule in differentiation, we know the derivative of a product ( u'v + uv' ). Let's consider ( u = g(x) ) and ( v = f(x) ), then:
[ g(x) = \int_a^x f(t)dt]
By the Fundamental Theorem, ( g'(x) = f(x)). Therefore:
[ \frac{d}{dx} \left( \frac{1}{2}[g(x)]^2 \right) = g(x)g'(x) = g(x)f(x) ]
This shows that ( \frac{1}{2}[g(x)]^2 ) is indeed a primitive of ( f(x)g(x) ).
Step 3
Conclusion
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Answer
Thus, the correct choice is option C: ( \frac{1}{2}[g(x)]^2 ).
