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Sagot :
To find the area of the region bounded by the graphs of the equations
[tex]\[ y = 7 + \sqrt[3]{x}, \quad x = 0, \quad x = 8, \quad y = 0 \][/tex]
we can follow these steps:
1. Understand the bounds and the function:
- The function [tex]\( y = 7 + \sqrt[3]{x} \)[/tex] represents a curve.
- The vertical lines [tex]\( x = 0 \)[/tex] and [tex]\( x = 8 \)[/tex] represent the lower and upper bounds of the integral.
- The line [tex]\( y = 0 \)[/tex] is the x-axis, which serves as the reference line for the area calculation.
2. Set up the integral:
To find the area under the curve [tex]\( y = 7 + \sqrt[3]{x} \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex], we need to integrate the function with respect to [tex]\( x \)[/tex] from 0 to 8.
The integral for the area [tex]\( A \)[/tex] is:
[tex]\[ A = \int_{0}^{8} \left( 7 + \sqrt[3]{x} \right) \, dx \][/tex]
3. Compute the integral:
Break the integral into two parts:
[tex]\[ A = \int_{0}^{8} 7 \, dx + \int_{0}^{8} \sqrt[3]{x} \, dx \][/tex]
First, evaluate the integral of the constant term [tex]\( 7 \)[/tex]:
[tex]\[ \int_{0}^{8} 7 \, dx = 7 \left[ x \right]_{0}^{8} = 7 \left( 8 - 0 \right) = 56 \][/tex]
Next, evaluate the integral of [tex]\( \sqrt[3]{x} \)[/tex]:
Recall that [tex]\( \sqrt[3]{x} \)[/tex] can be written as [tex]\( x^{1/3} \)[/tex]:
[tex]\[ \int_{0}^{8} x^{1/3} \, dx \][/tex]
To integrate [tex]\( x^{1/3} \)[/tex], use the power rule for integration:
[tex]\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \][/tex]
With [tex]\( n = 1/3 \)[/tex]:
[tex]\[ \int x^{1/3} \, dx = \frac{x^{(1/3)+1}}{(1/3)+1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \][/tex]
Evaluate this from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex]:
[tex]\[ \left[ \frac{3}{4} x^{4/3} \right]_{0}^{8} = \frac{3}{4} \left( 8^{4/3} - 0^{4/3} \right) \][/tex]
Simplify [tex]\( 8^{4/3} \)[/tex]:
Recall that [tex]\( 8^{4/3} = (2^3)^{4/3} = 2^4 = 16 \)[/tex]:
[tex]\[ \frac{3}{4} \left( 16 - 0 \right) = \frac{3}{4} \times 16 = 12 \][/tex]
Combine the results of the integrals:
[tex]\[ A = 56 + 12 = 68 \][/tex]
Thus, the area of the region bounded by the given graphs is:
[tex]\[ \boxed{68} \][/tex]
[tex]\[ y = 7 + \sqrt[3]{x}, \quad x = 0, \quad x = 8, \quad y = 0 \][/tex]
we can follow these steps:
1. Understand the bounds and the function:
- The function [tex]\( y = 7 + \sqrt[3]{x} \)[/tex] represents a curve.
- The vertical lines [tex]\( x = 0 \)[/tex] and [tex]\( x = 8 \)[/tex] represent the lower and upper bounds of the integral.
- The line [tex]\( y = 0 \)[/tex] is the x-axis, which serves as the reference line for the area calculation.
2. Set up the integral:
To find the area under the curve [tex]\( y = 7 + \sqrt[3]{x} \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex], we need to integrate the function with respect to [tex]\( x \)[/tex] from 0 to 8.
The integral for the area [tex]\( A \)[/tex] is:
[tex]\[ A = \int_{0}^{8} \left( 7 + \sqrt[3]{x} \right) \, dx \][/tex]
3. Compute the integral:
Break the integral into two parts:
[tex]\[ A = \int_{0}^{8} 7 \, dx + \int_{0}^{8} \sqrt[3]{x} \, dx \][/tex]
First, evaluate the integral of the constant term [tex]\( 7 \)[/tex]:
[tex]\[ \int_{0}^{8} 7 \, dx = 7 \left[ x \right]_{0}^{8} = 7 \left( 8 - 0 \right) = 56 \][/tex]
Next, evaluate the integral of [tex]\( \sqrt[3]{x} \)[/tex]:
Recall that [tex]\( \sqrt[3]{x} \)[/tex] can be written as [tex]\( x^{1/3} \)[/tex]:
[tex]\[ \int_{0}^{8} x^{1/3} \, dx \][/tex]
To integrate [tex]\( x^{1/3} \)[/tex], use the power rule for integration:
[tex]\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \][/tex]
With [tex]\( n = 1/3 \)[/tex]:
[tex]\[ \int x^{1/3} \, dx = \frac{x^{(1/3)+1}}{(1/3)+1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \][/tex]
Evaluate this from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex]:
[tex]\[ \left[ \frac{3}{4} x^{4/3} \right]_{0}^{8} = \frac{3}{4} \left( 8^{4/3} - 0^{4/3} \right) \][/tex]
Simplify [tex]\( 8^{4/3} \)[/tex]:
Recall that [tex]\( 8^{4/3} = (2^3)^{4/3} = 2^4 = 16 \)[/tex]:
[tex]\[ \frac{3}{4} \left( 16 - 0 \right) = \frac{3}{4} \times 16 = 12 \][/tex]
Combine the results of the integrals:
[tex]\[ A = 56 + 12 = 68 \][/tex]
Thus, the area of the region bounded by the given graphs is:
[tex]\[ \boxed{68} \][/tex]
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