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### Part (a): Area Between the Curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis
We are asked to find the area between [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis, bounded by the lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
1. Define the Function and Limits of Integration:
- The function is [tex]\( y = 2x^2 \)[/tex].
- The area is bounded by [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. Set Up the Integral:
The area under the curve from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex] is given by the definite integral:
[tex]\[ \text{Area} = \int_{1}^{3} 2x^2 \, dx \][/tex]
3. Evaluate the Integral:
- First, find the antiderivative of [tex]\( 2x^2 \)[/tex]:
[tex]\[ \int 2x^2 \, dx = \frac{2x^3}{3} + C \][/tex]
- Now, apply the limits of integration:
[tex]\[ \text{Area} = \left[ \frac{2x^3}{3} \right]_{1}^{3} \][/tex]
4. Compute the Value:
- Evaluate at [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{2 \cdot 3^3}{3} = \frac{2 \cdot 27}{3} = 18 \][/tex]
- Evaluate at [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{2 \cdot 1^3}{3} = \frac{2}{3} \][/tex]
- Find the difference:
[tex]\[ \text{Area} = 18 - \frac{2}{3} = \frac{54}{3} - \frac{2}{3} = \frac{52}{3} \approx 17.333333333333332 \][/tex]
So, the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis, from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex], is approximately [tex]\( 17.33 \)[/tex] square units.
### Part (b): Area Between the Curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis
For this part, we need to consider the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis, bounded by the lines at [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
1. Convert the Function to Solve for [tex]\( x \)[/tex]:
- The given function is [tex]\( y = 2x^2 \)[/tex].
- Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = \sqrt{\frac{y}{2}} \][/tex]
2. Define the Limits of Integration:
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex]:
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 2(1)^2 = 2 \)[/tex].
- At [tex]\( x = 3 \)[/tex], [tex]\( y = 2(3)^2 = 18 \)[/tex].
3. Set Up the Integral:
The area between the curve and the [tex]\( y \)[/tex]-axis from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex] is given by the definite integral with respect to [tex]\( y \)[/tex]:
[tex]\[ \text{Area} = \int_{2}^{18} \sqrt{\frac{y}{2}} \, dy \][/tex]
4. Evaluate the Integral:
- Simplify the integrand [tex]\( \sqrt{\frac{y}{2}} \)[/tex]:
[tex]\[ \sqrt{\frac{y}{2}} = \frac{1}{\sqrt{2}} \sqrt{y} = \frac{y^{1/2}}{\sqrt{2}} \][/tex]
- The integral becomes:
[tex]\[ \int_{2}^{18} \frac{y^{1/2}}{\sqrt{2}} \, dy \][/tex]
- Find the antiderivative of [tex]\( \frac{y^{1/2}}{\sqrt{2}} \)[/tex]:
[tex]\[ \int \frac{y^{1/2}}{\sqrt{2}} \, dy = \frac{1}{\sqrt{2}} \cdot \frac{2}{3} y^{3/2} = \frac{2}{3\sqrt{2}} y^{3/2} + C \][/tex]
- Apply the limits of integration:
[tex]\[ \text{Area} = \left[ \frac{2}{3\sqrt{2}} y^{3/2} \right]_{2}^{18} \][/tex]
5. Compute the Value:
- Evaluate at [tex]\( y = 18 \)[/tex]:
[tex]\[ \frac{2}{3\sqrt{2}} \cdot (18)^{3/2} \][/tex]
- Evaluate at [tex]\( y = 2 \)[/tex]:
[tex]\[ \frac{2}{3\sqrt{2}} \cdot (2)^{3/2} \][/tex]
- The difference gives the area:
[tex]\[ \text{Area} = 4 \][/tex]
So, the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis, from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex], is approximately [tex]\( 4.0 \)[/tex] square units.
Thus, the final areas are:
- Part (a): [tex]\( 17.333 \)[/tex] square units.
- Part (b): [tex]\( 4.0 \)[/tex] square units.
### Part (a): Area Between the Curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis
We are asked to find the area between [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis, bounded by the lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
1. Define the Function and Limits of Integration:
- The function is [tex]\( y = 2x^2 \)[/tex].
- The area is bounded by [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. Set Up the Integral:
The area under the curve from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex] is given by the definite integral:
[tex]\[ \text{Area} = \int_{1}^{3} 2x^2 \, dx \][/tex]
3. Evaluate the Integral:
- First, find the antiderivative of [tex]\( 2x^2 \)[/tex]:
[tex]\[ \int 2x^2 \, dx = \frac{2x^3}{3} + C \][/tex]
- Now, apply the limits of integration:
[tex]\[ \text{Area} = \left[ \frac{2x^3}{3} \right]_{1}^{3} \][/tex]
4. Compute the Value:
- Evaluate at [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{2 \cdot 3^3}{3} = \frac{2 \cdot 27}{3} = 18 \][/tex]
- Evaluate at [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{2 \cdot 1^3}{3} = \frac{2}{3} \][/tex]
- Find the difference:
[tex]\[ \text{Area} = 18 - \frac{2}{3} = \frac{54}{3} - \frac{2}{3} = \frac{52}{3} \approx 17.333333333333332 \][/tex]
So, the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( x \)[/tex]-axis, from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex], is approximately [tex]\( 17.33 \)[/tex] square units.
### Part (b): Area Between the Curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis
For this part, we need to consider the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis, bounded by the lines at [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
1. Convert the Function to Solve for [tex]\( x \)[/tex]:
- The given function is [tex]\( y = 2x^2 \)[/tex].
- Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = \sqrt{\frac{y}{2}} \][/tex]
2. Define the Limits of Integration:
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex]:
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 2(1)^2 = 2 \)[/tex].
- At [tex]\( x = 3 \)[/tex], [tex]\( y = 2(3)^2 = 18 \)[/tex].
3. Set Up the Integral:
The area between the curve and the [tex]\( y \)[/tex]-axis from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex] is given by the definite integral with respect to [tex]\( y \)[/tex]:
[tex]\[ \text{Area} = \int_{2}^{18} \sqrt{\frac{y}{2}} \, dy \][/tex]
4. Evaluate the Integral:
- Simplify the integrand [tex]\( \sqrt{\frac{y}{2}} \)[/tex]:
[tex]\[ \sqrt{\frac{y}{2}} = \frac{1}{\sqrt{2}} \sqrt{y} = \frac{y^{1/2}}{\sqrt{2}} \][/tex]
- The integral becomes:
[tex]\[ \int_{2}^{18} \frac{y^{1/2}}{\sqrt{2}} \, dy \][/tex]
- Find the antiderivative of [tex]\( \frac{y^{1/2}}{\sqrt{2}} \)[/tex]:
[tex]\[ \int \frac{y^{1/2}}{\sqrt{2}} \, dy = \frac{1}{\sqrt{2}} \cdot \frac{2}{3} y^{3/2} = \frac{2}{3\sqrt{2}} y^{3/2} + C \][/tex]
- Apply the limits of integration:
[tex]\[ \text{Area} = \left[ \frac{2}{3\sqrt{2}} y^{3/2} \right]_{2}^{18} \][/tex]
5. Compute the Value:
- Evaluate at [tex]\( y = 18 \)[/tex]:
[tex]\[ \frac{2}{3\sqrt{2}} \cdot (18)^{3/2} \][/tex]
- Evaluate at [tex]\( y = 2 \)[/tex]:
[tex]\[ \frac{2}{3\sqrt{2}} \cdot (2)^{3/2} \][/tex]
- The difference gives the area:
[tex]\[ \text{Area} = 4 \][/tex]
So, the area between the curve [tex]\( y = 2x^2 \)[/tex] and the [tex]\( y \)[/tex]-axis, from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex], is approximately [tex]\( 4.0 \)[/tex] square units.
Thus, the final areas are:
- Part (a): [tex]\( 17.333 \)[/tex] square units.
- Part (b): [tex]\( 4.0 \)[/tex] square units.
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