Integration
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| integration/ˌɪntɪˈɡreɪʃn/ | Tích phân |
| constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ | Hằng số tích phân |
| definite integral/ˈdefɪnət ˈɪntɪɡrəl/ | Tích phân xác định |
| improper integral/ɪmˈprɒpə ˈɪntɪɡrəl/ | An integral over an infinite interval or with an unbounded integrand, evaluated through suitable one-sided limits when they converge. |
| volume of revolution/ˈvɒljuːm ɒv ˌrevəˈluːʃn/ | Thể tích vật thể tròn xoay |
From a changing rate to a total
- A pump's water flow varies through the day. The total water delivered comes from accumulating flow rate over time; a fuel total similarly needs fuel-flow rate.
- Integration 积分 reverses differentiation. It can find a curve from its gradient, a signed total, a geometric area or a volume. Decide which quantity is requested.
Reverse the power and inner factor
- Add one to the power, divide by the new power, and divide by the coefficient of $x$. For rational $n\ne-1$ and $a\ne0$:
- Integrate sums and multiples term by term. For example, $\int(3x+2)^{1/2}\,dx=\dfrac29(3x+2)^{3/2}+C$. Differentiate the answer to check it. This power formula does not handle exponent $-1$.
Integrate $3(2x+1)^2-4(3x-2)^{-2}+5\sqrt{x}$ with respect to x, on an interval where x>0 and x≠2/3. Check by differentiation.
The result is $\tfrac12(2x+1)^3+\dfrac4{3(3x-2)}+\tfrac{10}3x^{3/2}+C$. Differentiation multiplies the first term by 2 and the reciprocal term by −3, recovering the original integrand.
Find each constant of integration 积分常数
- If $\dfrac{dy}{dx}=3x^2$ and the curve passes through $(1,5)$, integrate to $y=x^3+C$; then $5=1+C$, giving $C=4$.
- If you are given $y''$, integrate twice with two separate constants. Use a known gradient to fix the first constant and a known point to fix the second.
Given $y''=6x$ and a stationary point $(1,4)$, $y'=3x^2+A$. Since $y'(1)=0$, $A=-3$. Then $y=x^3-3x+B$; $4=1-3+B$ gives $B=6$.
Why does an indefinite integral include "+ C"?
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Differentiating a constant gives 0, so the original could have had any constant — hence + C.
If dy/dx = 4x and the curve passes through (1, 7), what is C in y = 2x² + C?
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y = 2x² + C. At (1, 7): 7 = 2(1)² + C → C = 5.
A curve has $dy/dx=\sqrt{2x+1}$ and passes through (4,10). Find its equation for x>−1/2.
$y=(2x+1)^{3/2}/3+C$. At x=4, $10=27/3+C$, so C=1 and $y=(2x+1)^{3/2}/3+1$.
For x>0 a curve has $y\prime\prime=12x-6x^{-4}$ and a stationary point (1,4). Find its equation.
$y\prime=6x^2+2x^{-3}+A$; $y\prime(1)=0$ gives A=−8. Integrating again gives $y=2x^3-x^{-2}-8x+B$. At (1,4), B=11. Hence $y=2x^3-1/x^2-8x+11$.
Definite integrals 定积分 are signed
- If $F'=f$, $\int_p^q f(x)\,dx=F(q)-F(p)$. There is no $+C$ in the final number: the constants cancel. Reversing the limits reverses its sign.
- With $p<q$, parts above the $x$-axis contribute positively and parts below negatively. Geometric area is never negative: split at every crossing and add the magnitudes.
Signed accumulation
Integral = ∫ f(x) dx
The definite integral measures signed accumulation: contributions below the x-axis are negative when the limits increase. Geometric area adds their magnitudes.
Evaluate the definite integral of 2x from 0 to 3 (∫₀³ 2x dx).
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∫2x dx = x², so [x²]₀³ = 9 − 0 = 9.
Evaluate ∫₀² 3x² dx.
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∫3x² dx = x³, so [x³]₀² = 8 − 0 = 8.
If f(x)<0 throughout p<x<q, with p<q, then its integral from p to q is negative.
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Area below the x-axis contributes negatively to the integral. Take the absolute value to find the actual area.
Evaluate $\int_1^3(2x-1)^{-2}dx$ and $\int_3^1(2x-1)^{-2}dx$. Explain why the signs differ although the integrand is positive.
An antiderivative is $-1/[2(2x-1)]$. The first integral is −1/10+1/2=2/5. The second is −2/5. A positive integrand gives a positive integral with increasing bounds; reversing the bounds changes the sign.
For $y=x^2-4$ on −3≤x≤3, find the definite integral and the total geometric area between the curve and x-axis.
An antiderivative is $F=x^3/3-4x$. The signed integral is F(3)−F(−3)=−6. Split at roots ±2. Each outside area is 7/3 and the middle area is 32/3. Total area is 46/3 square units.
Improper integrals 广义积分 as limits
- An improper integral has an infinite limit or an endpoint where the integrand is undefined. Replace that endpoint by a variable and take a limit; do not substitute infinity as a number.
- An unbounded integrand can still have a finite integral, as the first example shows. By contrast, $\int_0^1x^{-2}\,dx=\lim_{a\to0^+}(1/a-1)$ has no finite value.
Evaluate $\int_0^4x^{-1/2}dx$ and $\int_2^\infty6x^{-2}dx$ using limits. Determine whether $\int_0^1x^{-2}dx$ is finite.
The first is $\lim_{a\to0^+}(4-2\sqrt a)=4$. The second is $\lim_{b\to\infty}(3-6/b)=3$. The last is $\lim_{a\to0^+}(1/a-1)$, which grows without bound, so there is no finite integral.
Area between curves
- Find intersections to determine the limits. For vertical strips, integrate upper curve minus lower curve; if their order changes, split the interval.
- For $y=2x$ and $y=x^2$, intersections are $x=0,2$ and the line is above the curve between them. The area is $\int_0^2(2x-x^2)\,dx=[x^2-x^3/3]_0^2=4/3$.
Find the area enclosed by y=x² and y=x+2. Show the intersections and which curve is above.
$x^2=x+2$ gives x=−1,2. The line is above the parabola between them (for example, at x=0). Area $\int_{-1}^{2}(x+2-x^2)dx=9/2$ square units.
Volume of revolution 旋转体体积 about the x-axis
- A thin slice perpendicular to the axis has outer radius $R$ and inner radius $r$. Its area is $\pi(R^2-r^2)$. Integrate this cross-sectional area; a region touching the axis has $r=0$.
- Between $y=4-x^2$ and $y=1$, the limits are $\pm\sqrt3$. About the $x$-axis, $R=4-x^2,r=1$:
The volume of revolution of y = x about the x-axis from 0 to 2 is π∫₀² x² dx. Find the value (without π).
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∫₀² x² dx = [x³/3]₀² = 8/3 ≈ 2.67.
The region between y=6−x² and y=2 is rotated fully about the x-axis. Find its volume, explaining your radii.
Intersections give x=±2. Outer radius 6−x², inner radius 2. $V=\pi\int_{-2}^{2}[(6-x^2)^2-4]dx=\pi\int_{-2}^{2}(32-12x^2+x^4)dx=384\pi/5$ cubic units.
Rotate about the y-axis
- Use horizontal slices, write each radius as a distance in the $x$ direction, and integrate with respect to $y$: $V=\pi\int(R(y)^2-r(y)^2)\,dy$.
- For the region between $x=0$ and $x=\sqrt y$, $0\le y\le4$, the radius is $\sqrt y$. Thus $V=\pi\int_0^4y\,dy=8\pi$. Use $y$ limits, not the corresponding $x$ values.
The region 0≤y≤3, $1\le x\le\sqrt{y+1}$ is rotated fully about the y-axis. Find its volume.
Horizontal slices have outer radius √(y+1) and inner radius 1. $V=\pi\int_0^3[(y+1)-1]dy=9\pi/2$ cubic units.
Combine a tangent with an area
- On $y=2x-x^2$, the tangent at $(1,1)$ is $y=1$. Between $x=0$ and $x=1$, the area under the curve is $\int_0^1(2x-x^2)\,dx=2/3$.
- The rectangle beneath the tangent has area 1, so the gap between line and curve is $1-2/3=1/3$. In a larger problem, find the tangent and its intercepts before choosing the bounds and subtracting areas.
For $f(x)=4\sqrt{x+1}-x-4$ on 0≤x≤8, show its endpoints lie on the x-axis. Find the tangent at (8,0), its y-intercept, the area under the curve, and the area between tangent and curve on this interval.
$f(0)=f(8)=0$. $f\prime=2/\sqrt{x+1}-1$, so the endpoint gradient is −1/3. Tangent $y=(8-x)/3$, intercept (0,8/3). The area under the curve is $[\tfrac83(x+1)^{3/2}-x^2/2-4x]_0^8=16/3$. The tangent triangle has area 32/3, so the gap is 16/3. The line is above the curve since their difference is $\tfrac23(\sqrt{x+1}-3)^2\ge0$.
Choose the integral from the region
- A curve equation needs a constant; a definite integral needs limits. For geometric area, split where the signed contribution changes. For an improper integral, show the limiting calculation.
- For a volume, draw a slice perpendicular to the rotation axis. Identify outer and inner radii, square them separately, and use the matching integration variable and limits. State square units for area and cubic units for volume.
What is the integral of 6x from 1 to 3? If the antiderivative of 2x passes through (2, 7), what is the constant?
3 times 9 minus 3 times 1 is 24. 7 minus 4 is 3.
What is the definite integral of 6x from x=1 to x=3?
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27 minus 3 is 24.