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VICMath MethodsQuick questions

Unit 4

Quick questions on Integration by substitution: VCE Math Methods Unit 4

8short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is example 3. 1x\frac{1}{x}-style logarithmic?
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∫2x+3x2+3x+5 dx\int \frac{2x + 3}{x^2 + 3x + 5} \, dx.
What are option A. Change the limits?
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When u=g(x)u = g(x), the lower limit x=ax = a becomes u=g(a)u = g(a) and the upper limit x=bx = b becomes u=g(b)u = g(b). Evaluate the integral in uu directly between the new limits. No need to substitute back.
What is option B. Substitute back?
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Compute the antiderivative in uu, replace uu with g(x)g(x), then evaluate at the original xx-limits.
What are not substituting back in indefinite integrals?
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u55+c\frac{u^5}{5} + c is not a complete answer for an integral originally in xx; replace uu with g(x)g(x) before submitting.
What is constants getting lost?
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If the derivative of the inside is 2x2x but only xx is in the integrand, you must compensate with a factor of 12\frac{1}{2}. Forgetting halves the answer.
What is q1?
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Find ∫2x(x2+3)4 dx\displaystyle\int 2x(x^2 + 3)^4\,dx using u=x2+3u = x^2 + 3. [3 marks]
What is q2?
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Evaluate ∫01xex2 dx\displaystyle\int_0^1 x e^{x^2}\,dx. [3 marks]
What is q3?
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Find ∫cos⁡xsin⁡2x dx\displaystyle\int \cos x \sin^2 x\,dx. [3 marks]

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