Anti-differentiate by recognition
When the derivative of a function is supplied, use that relation to identify an anti-derivative directly. If the target is a constant multiple of the supplied derivative, keep the same multiple on the function
- Read the supplied relation A supplied derivative gives an anti-derivative without integrating its visible factors separately \(\displaystyle F'(x)=f(x)\Rightarrow\int f(x)\,dx=F(x)+C\)
- Carry a constant scale A constant multiplying the derivative also multiplies the recognised function \(\displaystyle g(x)=2F'(x)\Rightarrow\int g(x)\,dx=2F(x)+C\)
- Use one additive constant The complete family has one arbitrary constant for the whole anti-derivative \(\displaystyle \int\left(f(x)+h(x)\right)\,dx=F(x)+H(x)+C\)
- Differentiate to check Differentiate the proposed primitive and compare it with the target function \(\displaystyle \frac{d}{dx}\left(2F(x)+C\right)=2F'(x)\)
Worked example
Given the function and relation below, find the complete family of anti-derivatives of g
- \(\displaystyle F'(x)=\frac{d}{dx}\left(2\left(x^2+3\right)^{3/2}\right)=6x\sqrt{x^2+3}\)Differentiate the supplied function to make the derivative relation explicit
- \(\displaystyle g(x)=3F'(x)\)The target is three times the supplied derivative
- \(\displaystyle \int g(x)\,dx=3\int F'(x)\,dx=3F(x)+C\)Recognition returns F(x), while the scale factor 3 is preserved
- \(\displaystyle \int g(x)\,dx=6\left(x^2+3\right)^{3/2}+C\)Substitute the supplied expression for F(x) into the recognised family
Common mistakes
Difficulty levels
Questions about this topic
How is recognition different from applying the power rule?+
Recognition uses a supplied derivative relation to identify the original function. The visible expression is not split into separate factors for integration
Why is the same scale factor used after integration?+
Differentiation and anti-differentiation are linear, so a constant multiplying a derivative also multiplies its recognised function
How can a recognised anti-derivative be checked?+
Differentiate the proposed function and compare the result with the target integrand, including its scale factor
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Specialist Mathematics, chapter 5: Integral calculus