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- The definite integral of a function \(f(x)\) with respect to \(x\) with the interval from \(x=a \text{ to } x=b\) can be written as:
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\(\begin{aligned} \int_{a}^{b} f(x) \ dx &= [g(x)+c]_{a}^{b} \\ &=[g(b)+c]-[g(a)+c]\\ &=g(b)-g(a) \end{aligned}\) |
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| Characteristics of Definite Integrals |
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For the function \(f(x)\) and \(g(x)\), |
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\(\int_{a}^{b} f(x) \ dx = 0\) |
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\(\int_{a}^{b} f(x) \ dx = -\int_{a}^{b} f(x) \ dx\) |
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\(\int_{a}^{b} kf(x) \ dx = k\int_{a}^{b} f(x) \ dx\), where \(k\) is a constant |
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\(\int_{a}^{b} f(x) \ dx + \int_{b}^{c} f(x) \ dx = \int_{a}^{c} f(x) \ dx\), where \(a |
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\(\int_{a}^{b} [f(x) \pm g(x)] \ dx = \int_{a}^{b} f(x) \ dx \pm \int_{a}^{b}g(x) \ dx\) |
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- Area of a region between the curve and the \(x\)-axis :
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\(A = \int_{a}^{b} y \ dx\) |
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- For the value of the area bounded by the curve and the \(x\)-axis,
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If the region is below the \(x\)-axis, then the integral value is negative |
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If the region is above the \(x\)-axis, the the integral value is positive |
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The areas of both regions are positive |
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- Area between the curve and the \(y\)-axis :
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\(A= \int_{a}^{b} x \ dy\) |
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- For a region bounded by the curve and the \(y\)-axis,
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If the region is to the left of \(y\)-axis, then the integral value is negative |
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If the region is to the right of \(y\)-axis, then the integral value is positive |
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The areas of both regions are positive |
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- Area between the curve and a straight line:
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\(\begin{aligned} \text{The area of shaded region }&= \int_{a}^{b} g(x) \ dx -\int_{a}^{b} f(x) \ dx \\\\ &=\int_{a}^{b} [g(x)-f(x)] \ dx \end{aligned}\) |
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- To find the area, make sure that the function of the curve or line on the top is substracted to the function of the curve or line at the bottom
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| Example |
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The curves \(y=x^2\) and \(y= \sqrt[3] {x}\) intersect at points \((0,0) \text{ and } (1,1)\). |
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Find the area between the two curves. |
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Solution:
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\(\begin{aligned} \text{Area of the region }&= \int_{0}^{1} \sqrt[3] x \ dx - \int_{0}^{1} x^2 \ dx\\\\ &= \int_{0}^{1} (x^{\frac{1}{3}}-x^2) \ dx\\\\ &=\begin{bmatrix} \dfrac{3x^{\frac{4}{3}}}{4}-\dfrac{x^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\begin{bmatrix} \dfrac{3(1)^{\frac{4}{3}}}{4}-\dfrac{1^3}{3} \end{bmatrix}_{0}^{1} - \begin{bmatrix} \dfrac{3(0)^{\frac{4}{3}}}{4}-\dfrac{0^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\dfrac{5}{12} \text{ unit}^2 \end{aligned}\) |
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- The generated volume of a solid when an area is rotated through \(360^\text{o}\) about the \(x\)-axis or the \(y\)-axis
- The value of generated volume is always positive
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- Volume when a region is rotated through \(360^\text{o}\) about the \(x\)-axis :
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\(V= \int_{a}^{b} \pi y^2 \ dx\) |
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- Volume when a region is rotated through \(360^\text{o}\) about the \(y\)-axis :
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\(V = \int_{a}^{b} \pi x^2 \ dy\) |
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| Example |
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Based on the diagram above, the curve \(y=\dfrac{1}{4}x^2\) intersects the stright line \(y=x \text{ at } O(0,0) \text{ and }A(4,4)\). |
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Find the generated volume, in terms of \(\pi\), when the shaded region is revolved fully about the \(x\)-axis. |
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Solution: |
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Let \(V_{1} \) the volume generated by the straight line \(y=x\) and \(V_{2} \) be the volume generated by the curve \(y=\dfrac{1}{4}x^2\) from \(x=0 \text{ to }x=4\).
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| \(\begin{aligned} V_1 &= \int_{0}^{4} \pi (x)^2 \ dx\\\\ &=\pi \int_{0}^{4} (x)^2 \ dx\\\\ &= \pi \begin{bmatrix} \dfrac{x^3}{3} \end{bmatrix}_{0}^{4}\\\\ &= \pi \begin{bmatrix} \dfrac{4^3}{3} - \dfrac{0^3}{3} \end{bmatrix} \\\\ &=\dfrac{64}{3}\pi \text{ unit}^3 \end{aligned}\) |
\(\begin{aligned} V_2 &= \int_{0}^{4} \pi \begin{pmatrix} \dfrac{1}{4}x^2 \end{pmatrix}^2 \ dx\\\\ &=\pi \int_{0}^{4} \dfrac{1}{16}x^4 \ dx\\\\ &= \pi \begin{bmatrix} \dfrac{x^5}{16(5)} \end{bmatrix}_{0}^{4}\\\\ &= \pi \begin{bmatrix} \dfrac{4^5}{80} - \dfrac{0^5}{80} \end{bmatrix} \\\\ &=\dfrac{64}{5}\pi \text{ unit}^3 \end{aligned}\) |
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Thus, the generated volume |
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\(\begin{aligned} &= V_1 - V_2\\\\ &=\dfrac{64}{3} \pi -\dfrac{64}{5} \pi\\\\ &=8\dfrac{8}{15}\pi \text{ unit}^3 \end{aligned}\) |
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