Question
Find the area bounded by x−9x2−12x+35,y=0,x=8,andx=10.
Enter an exact answer.
Enter an exact answer.
Answers
GPT-4o mini
To find the area bounded by the curve \( y = x - 9x^2 - 12x + 35 \), the line \( y = 0 \), and the vertical lines \( x = 8 \) and \( x = 10 \), we will first simplify the expression for \( y \).
Starting with the equation:
\[
y = x - 9x^2 - 12x + 35
\]
we can simplify it:
\[
y = -9x^2 - 11x + 35.
\]
Next, we need to find the area between this curve and the x-axis from \( x = 8 \) to \( x = 10 \). The area can be found by integrating \( y \) from \( x = 8 \) to \( x = 10 \):
\[
\text{Area} = \int_{8}^{10} (-9x^2 - 11x + 35) \, dx.
\]
Now, we compute the integral:
1. First, calculate the antiderivative:
\[
\int (-9x^2 - 11x + 35) \, dx = -3x^3 - \frac{11}{2}x^2 + 35x + C.
\]
2. Now evaluate the definite integral from 8 to 10:
\[
\text{Area} = \left[-3x^3 - \frac{11}{2}x^2 + 35x\right]_{8}^{10}.
\]
Calculating at \( x = 10 \):
\[
-3(10)^3 - \frac{11}{2}(10)^2 + 35(10) = -3000 - 550 + 350 = -3200.
\]
Calculating at \( x = 8 \):
\[
-3(8)^3 - \frac{11}{2}(8)^2 + 35(8) = -1536 - 352 + 280 = -1608.
\]
Now substitute back into the integral:
\[
\text{Area} = \left(-3200 - (-1608)\right) = -3200 + 1608 = -1592.
\]
Since we are interested in the area which is a positive quantity, we take the absolute value:
\[
\text{Area} = 1592.
\]
Thus, the area bounded by the curve, the x-axis, and the lines \( x = 8 \) and \( x = 10 \) is:
\[
\boxed{1592}.
\]
Starting with the equation:
\[
y = x - 9x^2 - 12x + 35
\]
we can simplify it:
\[
y = -9x^2 - 11x + 35.
\]
Next, we need to find the area between this curve and the x-axis from \( x = 8 \) to \( x = 10 \). The area can be found by integrating \( y \) from \( x = 8 \) to \( x = 10 \):
\[
\text{Area} = \int_{8}^{10} (-9x^2 - 11x + 35) \, dx.
\]
Now, we compute the integral:
1. First, calculate the antiderivative:
\[
\int (-9x^2 - 11x + 35) \, dx = -3x^3 - \frac{11}{2}x^2 + 35x + C.
\]
2. Now evaluate the definite integral from 8 to 10:
\[
\text{Area} = \left[-3x^3 - \frac{11}{2}x^2 + 35x\right]_{8}^{10}.
\]
Calculating at \( x = 10 \):
\[
-3(10)^3 - \frac{11}{2}(10)^2 + 35(10) = -3000 - 550 + 350 = -3200.
\]
Calculating at \( x = 8 \):
\[
-3(8)^3 - \frac{11}{2}(8)^2 + 35(8) = -1536 - 352 + 280 = -1608.
\]
Now substitute back into the integral:
\[
\text{Area} = \left(-3200 - (-1608)\right) = -3200 + 1608 = -1592.
\]
Since we are interested in the area which is a positive quantity, we take the absolute value:
\[
\text{Area} = 1592.
\]
Thus, the area bounded by the curve, the x-axis, and the lines \( x = 8 \) and \( x = 10 \) is:
\[
\boxed{1592}.
\]