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

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

The formula for the area of the region \(A\) is given by:

\(A=\int^b_ay \, dx\)

For the value of the area bounded by the curve and the \(x\)-axis,

• If the region is below the \(x\)-axis, then the integral value is negative.
• If the region is above the \(x\)-axis, the the integral value is positive.
• The areas of both regions are positive.

The formula for the area of the region \(A\) is given by:

\(A=\int_a^b x \, dy\)

For a region bounded by the curve and the \(y\)-axis,

• If the region is to the left of \(y\)-axis, then the integral value is negative.
• If the region is to the right of \(y\)-axis, then the integral value is positive.
• The areas of both regions are positive.

The area of the shaded region can be illustrated as follows:

\(\begin{aligned} \text{The area of shaded region }&= \int_{a}^{b} g(x) \, dx -\int_{a}^{b} f(x) \, dx \\\\ &=\int_{a}^{b} [g(x)-f(x)] \, dx \end{aligned}\)

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.

• The generated volume of a solid when an area is rotated through \(360^\circ\) about the \(x\)-axis or \(y\)-axis.
• The value of generated volume is always positive.

The generated volume \(V\) when a region bounded by the curve \(y=f(x)\) enclosed by \(x=a\) and \(x=b\) is revolved through \(360^\circ\) about the \(x\)-axis is given by:

\(V=\int_a^b \pi y^2 \, dx\)

The generated volume \(V\) when a region bounded by the curve \(x=g(y)\) enclosed by \(y=a\) and \(y=b\) is revolved through \(360^\circ\) about the \(y\)-axis is given by:

\(V=\int_a^b \pi x^2 \, dy\)

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

\(\begin{aligned} \text{Area of the region }&= \int_{0}^{1} \sqrt[3] x \ dx - \int_{0}^{1} x^2 \ dx\\\\ &= \int_{0}^{1} (x^{\frac{1}{3}}-x^2) \ dx\\\\ &=\begin{bmatrix} \dfrac{3x^{\frac{4}{3}}}{4}-\dfrac{x^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\begin{bmatrix} \dfrac{3(1)^{\frac{4}{3}}}{4}-\dfrac{1^3}{3} \end{bmatrix}_{0}^{1} - \begin{bmatrix} \dfrac{3(0)^{\frac{4}{3}}}{4}-\dfrac{0^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\dfrac{5}{12} \text{ unit}^2 .\end{aligned}\)

Based on the diagram above, the curve \(y=\dfrac{1}{4}x^2\) intersects the stright line  \(y=x\) at \(O(0,0)\) and \(A(4,4)\). Find the generated volume, in terms of \(\pi\), when the shaded region is revolved fully about the \(x\)-axis.

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 \(x=0\) 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

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

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

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

The formula for the area of the region \(A\) is given by:

\(A=\int^b_ay \, dx\)

For the value of the area bounded by the curve and the \(x\)-axis,

• If the region is below the \(x\)-axis, then the integral value is negative.
• If the region is above the \(x\)-axis, the the integral value is positive.
• The areas of both regions are positive.

The formula for the area of the region \(A\) is given by:

\(A=\int_a^b x \, dy\)

For a region bounded by the curve and the \(y\)-axis,

• If the region is to the left of \(y\)-axis, then the integral value is negative.
• If the region is to the right of \(y\)-axis, then the integral value is positive.
• The areas of both regions are positive.

The area of the shaded region can be illustrated as follows:

\(\begin{aligned} \text{The area of shaded region }&= \int_{a}^{b} g(x) \, dx -\int_{a}^{b} f(x) \, dx \\\\ &=\int_{a}^{b} [g(x)-f(x)] \, dx \end{aligned}\)

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.

• The generated volume of a solid when an area is rotated through \(360^\circ\) about the \(x\)-axis or \(y\)-axis.
• The value of generated volume is always positive.

The generated volume \(V\) when a region bounded by the curve \(y=f(x)\) enclosed by \(x=a\) and \(x=b\) is revolved through \(360^\circ\) about the \(x\)-axis is given by:

\(V=\int_a^b \pi y^2 \, dx\)

The generated volume \(V\) when a region bounded by the curve \(x=g(y)\) enclosed by \(y=a\) and \(y=b\) is revolved through \(360^\circ\) about the \(y\)-axis is given by:

\(V=\int_a^b \pi x^2 \, dy\)

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

\(\begin{aligned} \text{Area of the region }&= \int_{0}^{1} \sqrt[3] x \ dx - \int_{0}^{1} x^2 \ dx\\\\ &= \int_{0}^{1} (x^{\frac{1}{3}}-x^2) \ dx\\\\ &=\begin{bmatrix} \dfrac{3x^{\frac{4}{3}}}{4}-\dfrac{x^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\begin{bmatrix} \dfrac{3(1)^{\frac{4}{3}}}{4}-\dfrac{1^3}{3} \end{bmatrix}_{0}^{1} - \begin{bmatrix} \dfrac{3(0)^{\frac{4}{3}}}{4}-\dfrac{0^3}{3} \end{bmatrix}_{0}^{1}\\\\ &=\dfrac{5}{12} \text{ unit}^2 .\end{aligned}\)

Based on the diagram above, the curve \(y=\dfrac{1}{4}x^2\) intersects the stright line  \(y=x\) at \(O(0,0)\) and \(A(4,4)\). Find the generated volume, in terms of \(\pi\), when the shaded region is revolved fully about the \(x\)-axis.

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 \(x=0\) 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

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