## Definite Integral

 3.3 Definite Integral

• 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:
 \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}

Characteristics of Definite Integrals

For the function $$f(x)$$ and $$g(x)$$,

 (a) $$\int_{a}^{b} f(x) \ dx = 0$$ (b) $$\int_{a}^{b} f(x) \ dx = -\int_{a}^{b} f(x) \ dx$$ (c) $$\int_{a}^{b} kf(x) \ dx = k\int_{a}^{b} f(x) \ dx$$, where $$k$$ is a constant (d) $$\int_{a}^{b} f(x) \ dx + \int_{b}^{c} f(x) \ dx = \int_{a}^{c} f(x) \ dx$$, where $$a (e) \(\int_{a}^{b} [f(x) \pm g(x)] \ dx = \int_{a}^{b} f(x) \ dx \pm \int_{a}^{b}g(x) \ dx$$

• Area of a region between the curve and the $$x$$-axis :
 $$A = \int_{a}^{b} y \ dx$$

• For the value of the area bounded by the curve and the $$x$$-axis,

 1. If the region is below the $$x$$-axis, then the integral value is negative 2. If the region is above the $$x$$-axis, the the integral value is positive 3. The areas of both regions are positive

• Area between the curve and the $$y$$-axis :
 $$A= \int_{a}^{b} x \ dy$$

• For a region bounded by the curve and the $$y$$-axis,

 1. If the region is to the left of $$y$$-axis, then the integral value is negative 2. If the region is to the right of $$y$$-axis, then the integral value is positive 3. The areas of both regions are positive

• Area between the curve and a straight line:
 \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}

• 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

 Example The curves $$y=x^2$$ and $$y= \sqrt[3] {x}$$ intersect at points $$(0,0) \text{ and } (1,1)$$. Find the area between the two curves. Solution: \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}

• 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

• Volume when a region is rotated through $$360^\text{o}$$ about the $$x$$-axis :
 $$V= \int_{a}^{b} \pi y^2 \ dx$$

• Volume when a region is rotated through $$360^\text{o}$$ about the $$y$$-axis :
 $$V = \int_{a}^{b} \pi x^2 \ dy$$

Example

Based on the diagram above, the curve $$y=\dfrac{1}{4}x^2$$ intersects the stright line  $$y=x \text{ at } O(0,0) \text{ and }A(4,4)$$.

Find the generated volume, in terms of $$\pi$$, when the shaded region is revolved fully about the $$x$$-axis.

Solution:

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 $$x=0 \text{ to }x=4$$.

 \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}

Thus, the generated volume

\begin{aligned} &= V_1 - V_2\\\\ &=\dfrac{64}{3} \pi -\dfrac{64}{5} \pi\\\\ &=8\dfrac{8}{15}\pi \text{ unit}^3 \end{aligned}

## Definite Integral

 3.3 Definite Integral

• 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:
 \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}

Characteristics of Definite Integrals

For the function $$f(x)$$ and $$g(x)$$,

 (a) $$\int_{a}^{b} f(x) \ dx = 0$$ (b) $$\int_{a}^{b} f(x) \ dx = -\int_{a}^{b} f(x) \ dx$$ (c) $$\int_{a}^{b} kf(x) \ dx = k\int_{a}^{b} f(x) \ dx$$, where $$k$$ is a constant (d) $$\int_{a}^{b} f(x) \ dx + \int_{b}^{c} f(x) \ dx = \int_{a}^{c} f(x) \ dx$$, where $$a (e) \(\int_{a}^{b} [f(x) \pm g(x)] \ dx = \int_{a}^{b} f(x) \ dx \pm \int_{a}^{b}g(x) \ dx$$

• Area of a region between the curve and the $$x$$-axis :
 $$A = \int_{a}^{b} y \ dx$$

• For the value of the area bounded by the curve and the $$x$$-axis,

 1. If the region is below the $$x$$-axis, then the integral value is negative 2. If the region is above the $$x$$-axis, the the integral value is positive 3. The areas of both regions are positive

• Area between the curve and the $$y$$-axis :
 $$A= \int_{a}^{b} x \ dy$$

• For a region bounded by the curve and the $$y$$-axis,

 1. If the region is to the left of $$y$$-axis, then the integral value is negative 2. If the region is to the right of $$y$$-axis, then the integral value is positive 3. The areas of both regions are positive

• Area between the curve and a straight line:
 \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}

• 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

 Example The curves $$y=x^2$$ and $$y= \sqrt[3] {x}$$ intersect at points $$(0,0) \text{ and } (1,1)$$. Find the area between the two curves. Solution: \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}

• 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

• Volume when a region is rotated through $$360^\text{o}$$ about the $$x$$-axis :
 $$V= \int_{a}^{b} \pi y^2 \ dx$$

• Volume when a region is rotated through $$360^\text{o}$$ about the $$y$$-axis :
 $$V = \int_{a}^{b} \pi x^2 \ dy$$

Example

Based on the diagram above, the curve $$y=\dfrac{1}{4}x^2$$ intersects the stright line  $$y=x \text{ at } O(0,0) \text{ and }A(4,4)$$.

Find the generated volume, in terms of $$\pi$$, when the shaded region is revolved fully about the $$x$$-axis.

Solution:

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 $$x=0 \text{ to }x=4$$.

 \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}

Thus, the generated volume

\begin{aligned} &= V_1 - V_2\\\\ &=\dfrac{64}{3} \pi -\dfrac{64}{5} \pi\\\\ &=8\dfrac{8}{15}\pi \text{ unit}^3 \end{aligned}