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To find the area bounded by the curve [tex]\(y=2x^2+x+1\)[/tex], the x-axis, and the vertical lines [tex]\(x=1\)[/tex] and [tex]\(x=3\)[/tex], we need to evaluate the definite integral of the function [tex]\(y=2x^2+x+1\)[/tex] with respect to [tex]\(x\)[/tex] from 1 to 3.
Let's break down the steps:
1. Define the function to be integrated: The function given is [tex]\( f(x) = 2x^2 + x + 1 \)[/tex].
2. Set up the definite integral: To find the bounded area, we need to integrate [tex]\( f(x) \)[/tex] from 1 to 3.
[tex]\[ \text{Area} = \int_{1}^{3} (2x^2 + x + 1) \, dx \][/tex]
3. Integrate the function: We will find the antiderivative of [tex]\( f(x) \)[/tex]:
[tex]\[ \int (2x^2 + x + 1) \, dx \][/tex]
- The antiderivative of [tex]\(2x^2\)[/tex] is [tex]\(\frac{2x^3}{3}\)[/tex].
- The antiderivative of [tex]\(x\)[/tex] is [tex]\(\frac{x^2}{2}\)[/tex].
- The antiderivative of [tex]\(1\)[/tex] is [tex]\(x\)[/tex].
Therefore,
[tex]\[ \int (2x^2 + x + 1) \, dx = \frac{2x^3}{3} + \frac{x^2}{2} + x + C \][/tex]
4. Evaluate the definite integral: We need to find the value of the antiderivative at the upper and lower limits and subtract the two.
Evaluating the antiderivative at [tex]\(x = 3\)[/tex]:
[tex]\[ \left[ \frac{2(3)^3}{3} + \frac{(3)^2}{2} + 3 \right] = \left[ \frac{54}{3} + \frac{9}{2} + 3 \right] \][/tex]
[tex]\[ = 18 + 4.5 + 3 = 25.5 \][/tex]
Evaluating the antiderivative at [tex]\(x = 1\)[/tex]:
[tex]\[ \left[ \frac{2(1)^3}{3} + \frac{(1)^2}{2} + 1 \right] = \left[ \frac{2}{3} + \frac{1}{2} + 1 \right] \][/tex]
[tex]\[ = \frac{2}{3} + \frac{1}{2} + 1 = \frac{4}{6} + \frac{3}{6} + 1 = \frac{7}{6} + 1 = \frac{13}{6} \approx 2.1667 \][/tex]
Finally, subtract these two results:
[tex]\[ \left( 25.5 \right) - \left( \frac{13}{6} \right) \][/tex]
Converting [tex]\(\frac{13}{6}\)[/tex] to a decimal:
[tex]\[ \frac{13}{6} \approx 2.1667 \][/tex]
So the area is:
[tex]\[ 25.5 - 2.1667 \approx 23.3333 \][/tex]
Thus, the area bounded by the curve [tex]\( y = 2x^2 + x + 1 \)[/tex], the x-axis, and the vertical lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex] is approximately [tex]\( 23.33 \)[/tex] square units.
Let's break down the steps:
1. Define the function to be integrated: The function given is [tex]\( f(x) = 2x^2 + x + 1 \)[/tex].
2. Set up the definite integral: To find the bounded area, we need to integrate [tex]\( f(x) \)[/tex] from 1 to 3.
[tex]\[ \text{Area} = \int_{1}^{3} (2x^2 + x + 1) \, dx \][/tex]
3. Integrate the function: We will find the antiderivative of [tex]\( f(x) \)[/tex]:
[tex]\[ \int (2x^2 + x + 1) \, dx \][/tex]
- The antiderivative of [tex]\(2x^2\)[/tex] is [tex]\(\frac{2x^3}{3}\)[/tex].
- The antiderivative of [tex]\(x\)[/tex] is [tex]\(\frac{x^2}{2}\)[/tex].
- The antiderivative of [tex]\(1\)[/tex] is [tex]\(x\)[/tex].
Therefore,
[tex]\[ \int (2x^2 + x + 1) \, dx = \frac{2x^3}{3} + \frac{x^2}{2} + x + C \][/tex]
4. Evaluate the definite integral: We need to find the value of the antiderivative at the upper and lower limits and subtract the two.
Evaluating the antiderivative at [tex]\(x = 3\)[/tex]:
[tex]\[ \left[ \frac{2(3)^3}{3} + \frac{(3)^2}{2} + 3 \right] = \left[ \frac{54}{3} + \frac{9}{2} + 3 \right] \][/tex]
[tex]\[ = 18 + 4.5 + 3 = 25.5 \][/tex]
Evaluating the antiderivative at [tex]\(x = 1\)[/tex]:
[tex]\[ \left[ \frac{2(1)^3}{3} + \frac{(1)^2}{2} + 1 \right] = \left[ \frac{2}{3} + \frac{1}{2} + 1 \right] \][/tex]
[tex]\[ = \frac{2}{3} + \frac{1}{2} + 1 = \frac{4}{6} + \frac{3}{6} + 1 = \frac{7}{6} + 1 = \frac{13}{6} \approx 2.1667 \][/tex]
Finally, subtract these two results:
[tex]\[ \left( 25.5 \right) - \left( \frac{13}{6} \right) \][/tex]
Converting [tex]\(\frac{13}{6}\)[/tex] to a decimal:
[tex]\[ \frac{13}{6} \approx 2.1667 \][/tex]
So the area is:
[tex]\[ 25.5 - 2.1667 \approx 23.3333 \][/tex]
Thus, the area bounded by the curve [tex]\( y = 2x^2 + x + 1 \)[/tex], the x-axis, and the vertical lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex] is approximately [tex]\( 23.33 \)[/tex] square units.
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