Answer :
To find the area between the graphs of the equations [tex]\( y = 3x^2 \)[/tex] and [tex]\( y = 4x \)[/tex], we will follow these steps:
1. Find the intersection points of the graphs:
To determine where the two curves intersect, solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 = 4x \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 4x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(3x - 4) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = 0 \quad \text{and} \quad x = \frac{4}{3} \][/tex]
2. Set up the definite integrals:
The area between the curves from [tex]\( x = 0 \)[/tex] to [tex]\( x = \frac{4}{3} \)[/tex] can be determined by finding the integral of the upper function minus the integral of the lower function:
[tex]\[ \text{Area} = \int_{0}^{\frac{4}{3}} (4x - 3x^2) \, dx \][/tex]
3. Evaluate the definite integrals:
First, integrate each function separately:
For [tex]\( 4x \)[/tex]:
[tex]\[ \int 4x \, dx = 2x^2 \][/tex]
For [tex]\( 3x^2 \)[/tex]:
[tex]\[ \int 3x^2 \, dx = x^3 \][/tex]
Now, apply the limits of integration ([tex]\( 0 \)[/tex] to [tex]\( \frac{4}{3} \)[/tex]):
[tex]\[ \left[ 2x^2 \right]_0^{\frac{4}{3}} = 2 \left(\frac{4}{3}\right)^2 - 2(0)^2 = 2 \cdot \frac{16}{9} = \frac{32}{9} \][/tex]
[tex]\[ \left[ x^3 \right]_0^{\frac{4}{3}} = \left(\frac{4}{3}\right)^3 - (0)^3 = \frac{64}{27} \][/tex]
4. Find the difference between the areas:
The area between the graphs is given by the difference of these integrals:
[tex]\[ \text{Area} = \frac{32}{9} - \frac{64}{27} \][/tex]
To subtract these fractions, find a common denominator (in this case, it's 27):
[tex]\[ \frac{32}{9} = \frac{32 \cdot 3}{9 \cdot 3} = \frac{96}{27} \][/tex]
Now subtract the fractions:
[tex]\[ \frac{96}{27} - \frac{64}{27} = \frac{32}{27} \][/tex]
5. Round the result to two decimal places:
Convert [tex]\(\frac{32}{27}\)[/tex] to a decimal:
[tex]\[ \frac{32}{27} \approx 1.185 \approx 1.19 \][/tex]
So, the area between [tex]\( y = 3x^2 \)[/tex] and [tex]\( y = 4x \)[/tex], rounded to two decimal places, is [tex]\( 1.19 \)[/tex] square units.
1. Find the intersection points of the graphs:
To determine where the two curves intersect, solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 = 4x \][/tex]
Rearrange the equation:
[tex]\[ 3x^2 - 4x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(3x - 4) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = 0 \quad \text{and} \quad x = \frac{4}{3} \][/tex]
2. Set up the definite integrals:
The area between the curves from [tex]\( x = 0 \)[/tex] to [tex]\( x = \frac{4}{3} \)[/tex] can be determined by finding the integral of the upper function minus the integral of the lower function:
[tex]\[ \text{Area} = \int_{0}^{\frac{4}{3}} (4x - 3x^2) \, dx \][/tex]
3. Evaluate the definite integrals:
First, integrate each function separately:
For [tex]\( 4x \)[/tex]:
[tex]\[ \int 4x \, dx = 2x^2 \][/tex]
For [tex]\( 3x^2 \)[/tex]:
[tex]\[ \int 3x^2 \, dx = x^3 \][/tex]
Now, apply the limits of integration ([tex]\( 0 \)[/tex] to [tex]\( \frac{4}{3} \)[/tex]):
[tex]\[ \left[ 2x^2 \right]_0^{\frac{4}{3}} = 2 \left(\frac{4}{3}\right)^2 - 2(0)^2 = 2 \cdot \frac{16}{9} = \frac{32}{9} \][/tex]
[tex]\[ \left[ x^3 \right]_0^{\frac{4}{3}} = \left(\frac{4}{3}\right)^3 - (0)^3 = \frac{64}{27} \][/tex]
4. Find the difference between the areas:
The area between the graphs is given by the difference of these integrals:
[tex]\[ \text{Area} = \frac{32}{9} - \frac{64}{27} \][/tex]
To subtract these fractions, find a common denominator (in this case, it's 27):
[tex]\[ \frac{32}{9} = \frac{32 \cdot 3}{9 \cdot 3} = \frac{96}{27} \][/tex]
Now subtract the fractions:
[tex]\[ \frac{96}{27} - \frac{64}{27} = \frac{32}{27} \][/tex]
5. Round the result to two decimal places:
Convert [tex]\(\frac{32}{27}\)[/tex] to a decimal:
[tex]\[ \frac{32}{27} \approx 1.185 \approx 1.19 \][/tex]
So, the area between [tex]\( y = 3x^2 \)[/tex] and [tex]\( y = 4x \)[/tex], rounded to two decimal places, is [tex]\( 1.19 \)[/tex] square units.
