Integration
Introduced| English | Meaning |
|---|---|
| integration/ˌɪntɪˈɡreɪʃn/ | Finding an integral, such as an antiderivative or the accumulated value over an interval. |
| definite integral/ˈdefɪnət ˈɪntɪɡrəl/ | The limit of sums of function values multiplied by interval widths, representing signed area under a curve over an interval. |
| fundamental theorem of calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ | The theorem linking differentiation and integration: accumulation differentiates to its continuous integrand, and definite integrals can be evaluated using antiderivatives. |
| constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ | An arbitrary constant included in an indefinite integral to represent the family of antiderivatives. |
| integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ | integration by substitution |
Signed accumulation can cancel
- Integration 积分 can recover an antiderivative or calculate signed accumulation. A definite integral is not automatically a nonnegative geometric area.
- For a continuous integrand, the fundamental theorem of calculus 微积分基本定理 connects a definite integral to an antiderivative evaluated at its bounds.
Keep the antiderivative family
- An indefinite integral needs a constant of integration 积分常数 because differentiating any constant gives zero. The power formula is $\int x^n dx=x^{n+1}/(n+1)+C$ for $n\ne-1$ on a valid domain.
- The reciprocal exception is $\int1/x\,dx=\ln|x|+C$ on an interval not crossing zero. A positive or negative domain needs attention before choosing a logarithm expression.
Integrate x³ with respect to x. Include the constant.
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Add one to the power and divide by the new power, then add c because any constant differentiates to zero.
A definite integral needs a constant of integration.
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The constant cancels in F(b) − F(a). Only an indefinite integral needs it — and there it is compulsory.
In one English sentence, explain why an indefinite integral needs "+ c".
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Example: "Any constant differentiates to zero, so the original constant cannot be recovered from the derivative."
Evaluate upper minus lower
- A definite integral 定积分 gives $\int_a^bf(x)dx=F(b)-F(a)$ when the antiderivative method applies. Substitute both bounds into the same F.
- C cancels between those evaluations. For geometric area, split where the curve changes sign or where upper and lower curves exchange order.
The area under the curve
$\int_a^b f(x)\,dx$
A definite integral is the signed area between the curve and the axis.
Evaluate the integral of x² from 0 to 3.
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[x³/3] from 0 to 3 = 27/3 − 0 = 9.
Zero integral, positive area. For $y=x$ from negative 1 to 1, $I=[x^2/2]_{-1}^1=0$. Geometric area is $A=-\int_{-1}^0x\,dx+\int_0^1x\,dx=1$. Each triangular region has area one half; their signed contributions cancel.
Substitute both the function and differential
- Integration by substitution 换元积分法 reverses the chain rule. With $u=x^2+1$, $du=2x\,dx$, so $\int2x(x^2+1)^5dx=\int u^5du$.
- For a definite integral, transform its bounds to u or return to x before evaluating original bounds. Do not combine u expressions with x bounds.
Which substitution simplifies the integral of 2x(x² + 1)⁵?
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Its derivative 2x is already present, so du = 2x dx and the integral becomes u⁵ du.
Differentiate an antiderivative to check it. Also check the domain, integration constant and bounds; differentiation alone does not prove that a definite integral or geometric area was evaluated correctly. Sheet 3.2 uses velocity to distinguish displacement, total distance and final position.
How do you check an integration answer reliably?
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Differentiate a proposed antiderivative and compare it with the original integrand on its domain. Also check the constant, bounds and sign or geometric-area interpretation; differentiation alone does not validate those steps.
An antiderivative valid on an interval cannot automatically justify evaluating across a singularity. Check the integrand's domain before applying upper-minus-lower. An initial position must be added when an integral gives only a displacement.