Elementary antiderivatives, coefficients and domains
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| antiderivative/ˌæntɪdɪˈrɪvətɪv/ | nguyên hàm |
How must integration undo that factor?
- Differentiation multiplies an exponential or trigonometric argument by its coefficient.
- This lesson studies antiderivative 原函数: A function whose derivative is the specified integrand on the stated interval.
The method
- For real powers on an allowed differentiable domain, integrate x^n as x^(n+1)/(n+1)+C when n≠−1.
- Integrate 1/x as ln|x|+C on an interval excluding zero.
- For k≠0, integrals of e^(kx), sin(kx), cos(kx) are e^(kx)/k, −cos(kx)/k, sin(kx)/k respectively, plus constants.
- Apply linearity to sums and constant multiples; verify by differentiation.
Why is the integral of 1/x not x⁰/0 or any power rule case?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The power rule divides by n+1, which fails exactly at n=-1. That gap is filled by the logarithm: (ln|x|)′=1/x on either side of zero, with |x| covering negative inputs.
A worked case
For f(x)=6x²−4e^(2x)+3cos(3x)−2sin(2x)+5/x with x>0, F=2x³−2e^(2x)+sin(3x)+cos(2x)+5ln x+C. Each derivative recovers its matching term. On x<0, integrate 5/x as 5ln(−x), or 5ln|x|; constants can differ on disconnected intervals. The integral of 3cos(3x) from 0 to π/6 is sin(π/2)−sin 0=1. For F′=6x² and F(1)=7, F=2x³+5. If k=0, e^(kx)=1 and cos(kx)=1 integrate to x+C, while sin(kx)=0 integrates to C; never divide by k=0.
Elementary antiderivatives, coefficients and domains
For real powers on an allowed differentiable domain, integrate x^n as x^(n+1)/(n+1)+C when n≠−1
Check the hypothesis, endpoints and coefficients behind each integral calculation.
For F′=6x² and F(1)=7, find the integration constant.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
7=2×1³+C gives C=5.
Where it goes wrong
- The n=−1 power is a separate logarithmic case, not division by zero in the power formula.
- Keep the sign in the sine antiderivative and divide by the argument coefficient. ln x alone does not cover negative x.
- An indefinite integral needs a constant; a definite integral has a numerical value when valid limits are supplied.
- A real fractional power may restrict the domain.
Not true: Integrating cos(kx) gives k sin(kx), including k=0. ∫cos(kx) dx=(1/k)sin(kx)+c for k≠0; when k=0 the integrand is 1 and the integral is x+c.
Find the worked definite integral of 3cos(3x).
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The antiderivative sin(3x) gives sin(π/2)−sin 0=1.
∫cos(kx) dx=(1/k)sin(kx)+c for k≠0.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. ∫cos(kx) dx=(1/k)sin(kx)+c for k≠0; when k=0 the integrand is 1 and the integral is x+c.
Applying it
- Differentiate term by term to check coefficients and signs, and state the interval before using logarithms or fractional powers.
- The H2 named elementary forms are powers, e^(kx), 1/x, sin(kx) and cos(kx).
- Other forms require separate justification and are not asserted as additional compulsory H2 forms.
Evaluate the derivative of 5ln|x| at x=−2.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The derivative is 5/x on either allowed interval, giving −5/2.
Match each integrand to its antiderivative.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Verify every antiderivative by differentiating back; sine carries the minus sign, and k=0 degenerates to constants needing separate treatment.
In your course
- AQA A-level Mathematics (7357), specification section H.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Antiderivative: A function whose derivative is the specified integrand on the stated interval.
- For real powers on an allowed differentiable domain, integrate x^n as x^(n+1)/(n+1)+C when n≠−1.
- Watch for: the n=−1 power is a separate logarithmic case, not division by zero in the power formula.