AP Calculus AB Exam Question:
Consider the function f(x) given by [ f(x) = x^3 - 4x^2 + 5x - 2 ]. Let R be the region bounded by the x-axis and the curve f(x) in the interval [ 0 \leq x \leq 2 ].
a) Find the exact area of region R.
b) The line y = -2 intersects the curve f(x) at two points. Find the coordinates of these points.
c) Calculate the exact volume of the solid generated when the region R is revolved about the x-axis.
Answer:
a) To find the exact area of region R, we need to compute the definite integral of f(x) from 0 to 2:
∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x \int_0^2 (x^3 - 4x^2 + 5x - 2) dx ∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x First, we find the antiderivative F(x) of f(x):
F ( x ) = x 4 4 − 4 x 3 3 + 5 x 2 2 − 2 x + C F(x) = \frac{x^4}{4} - \frac{4x^3}{3} + \frac{5x^2}{2} - 2x + C F ( x ) = 4 x 4 − 3 4 x 3 + 2 5 x 2 − 2 x + C Applying the Fundamental Theorem of Calculus, we have:
∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = F ( 2 ) − F ( 0 ) \int_0^2 (x^3 - 4x^2 + 5x - 2) dx = F(2) - F(0) ∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = F ( 2 ) − F ( 0 ) Plugging in the values, we get:
∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = ( 2 4 4 − 4 ( 2 3 ) 3 + 5 ( 2 2 ) 2 − 2 ( 2 ) ) − ( 0 4 4 − 4 ( 0 3 ) 3 + 5 ( 0 2 ) 2 − 2 ( 0 ) ) \int_0^2 (x^3 - 4x^2 + 5x - 2) dx = (\frac{2^4}{4} - \frac{4(2^3)}{3} + \frac{5(2^2)}{2} - 2(2)) - (\frac{0^4}{4} - \frac{4(0^3)}{3} + \frac{5(0^2)}{2} - 2(0)) ∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = ( 4 2 4 − 3 4 ( 2 3 ) + 2 5 ( 2 2 ) − 2 ( 2 )) − ( 4 0 4 − 3 4 ( 0 3 ) + 2 5 ( 0 2 ) − 2 ( 0 )) Simplifying further:
∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = ( 16 4 − 32 3 + 20 2 − 4 ) − ( 0 − 0 + 0 − 0 ) = 4 − 32 3 + 10 − 4 = 14 3 \int_0^2 (x^3 - 4x^2 + 5x - 2) dx = (\frac{16}{4} - \frac{32}{3} + \frac{20}{2} - 4) - (0 - 0 + 0 - 0) = 4 - \frac{32}{3} + 10 - 4 = \frac{14}{3} ∫ 0 2 ( x 3 − 4 x 2 + 5 x − 2 ) d x = ( 4 16 − 3 32 + 2 20 − 4 ) − ( 0 − 0 + 0 − 0 ) = 4 − 3 32 + 10 − 4 = 3 14 Therefore, the exact area of region R is [ \frac{14}{3} ].
b) To find the coordinates of the points where the line y = -2 intersects the curve f(x), we set f(x) equal to -2:
x 3 − 4 x 2 + 5 x − 2 = − 2 x^3 - 4x^2 + 5x - 2 = -2 x 3 − 4 x 2 + 5 x − 2 = − 2 Simplifying the equation:
x 3 − 4 x 2 + 5 x = 0 x^3 - 4x^2 + 5x = 0 x 3 − 4 x 2 + 5 x = 0 Factoring out an x:
x ( x 2 − 4 x + 5 ) = 0 x(x^2 - 4x + 5) = 0 x ( x 2 − 4 x + 5 ) = 0 This equation is satisfied when x = 0 and when the quadratic equation x^2 - 4x + 5 = 0 has real roots. Using the quadratic formula, we find:
x = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} x = 2 ( 1 ) − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( 5 ) x = 4 ± 16 − 20 2 x = \frac{4 \pm \sqrt{16 - 20}}{2} x = 2 4 ± 16 − 20 x = 4 ± − 4 2 x = \frac{4 \pm \sqrt{-4}}{2} x = 2 4 ± − 4 x = 4 ± 2 i 2 x = \frac{4 \pm 2i}{2} x = 2 4 ± 2 i Since we are looking for real roots, there are no intersection points between y = -2 and the curve f(x). Hence, there are no coordinates to be found.
c) To find the exact volume of the solid generated when the region R is revolved about the x-axis, we use the disc method. The volume V is given by:
V = π ∫ 0 2 [ f ( x ) ] 2 d x V = \pi \int_0^2 [f(x)]^2 dx V = π ∫ 0 2 [ f ( x ) ] 2 d x We substitute f(x) into the equation:
V = π ∫ 0 2 [ ( x 3 − 4 x 2 + 5 x − 2 ) 2 ] d x V = \pi \int_0^2 [(x^3 - 4x^2 + 5x - 2)^2] dx V = π ∫ 0 2 [( x 3 − 4 x 2 + 5 x − 2 ) 2 ] d x Expanding and simplifying the equation:
V = π ∫ 0 2 ( x 6 − 8 x 5 + 24 x 4 − 32 x 3 + 20 x 2 − 4 x + 4 ) d x V = \pi \int_0^2 (x^6 - 8x^5 + 24x^4 - 32x^3 + 20x^2 - 4x + 4) dx V = π ∫ 0 2 ( x 6 − 8 x 5 + 24 x 4 − 32 x 3 + 20 x 2 − 4 x + 4 ) d x V = π [ x 7 7 − 8 x 6 6 + 24 x 5 5 − 32 x 4 4 + 20 x 3 3 − 4 x 2 2 + 4 x ] 0 2 V = \pi \left[ \frac{x^7}{7} - \frac{8x^6}{6} + \frac{24x^5}{5} - \frac{32x^4}{4} + \frac{20x^3}{3} - \frac{4x^2}{2} + 4x \right]_0^2 V = π [ 7 x 7 − 6 8 x 6 + 5 24 x 5 − 4 32 x 4 + 3 20 x 3 − 2 4 x 2 + 4 x ] 0 2 Applying the limits:
V = π [ 2 7 7 − 8 ( 2 6 ) 6 + 24 ( 2 5 ) 5 − 32 ( 2 4 ) 4 + 20 ( 2 3 ) 3 − 4 ( 2 2 ) 2 + 4 ( 2 ) ] − π [ 0 7 7 − 8 ( 0 6 ) 6 + 24 ( 0 5 ) 5 − 32 ( 0 4 ) 4 + 20 ( 0 3 ) 3 − 4 ( 0 2 ) 2 + 4 ( 0 ) ] V = \pi \left[ \frac{2^7}{7} - \frac{8(2^6)}{6} + \frac{24(2^5)}{5} - \frac{32(2^4)}{4} + \frac{20(2^3)}{3} - \frac{4(2^2)}{2} + 4(2) \right] - \pi \left[ \frac{0^7}{7} - \frac{8(0^6)}{6} + \frac{24(0^5)}{5} - \frac{32(0^4)}{4} + \frac{20(0^3)}{3} - \frac{4(0^2)}{2} + 4(0) \right] V = π [ 7 2 7 − 6 8 ( 2 6 ) + 5 24 ( 2 5 ) − 4 32 ( 2 4 ) + 3 20 ( 2 3 ) − 2 4 ( 2 2 ) + 4 ( 2 ) ] − π [ 7 0 7 − 6 8 ( 0 6 ) + 5 24 ( 0 5 ) − 4 32 ( 0 4 ) + 3 20 ( 0 3 ) − 2 4 ( 0 2 ) + 4 ( 0 ) ] Simplifying further:
V = π [ 128 7 − 128 3 + 192 5 − 32 + 160 3 − 8 + 8 ] − π [ 0 − 0 + 0 − 0 + 0 − 0 + 0 ] V = \pi \left[ \frac{128}{7} - \frac{128}{3} + \frac{192}{5} - 32 + \frac{160}{3} - 8 + 8 \right] - \pi \left[ 0 - 0 + 0 - 0 + 0 - 0 + 0 \right] V = π [ 7 128 − 3 128 + 5 192 − 32 + 3 160 − 8 + 8 ] − π [ 0 − 0 + 0 − 0 + 0 − 0 + 0 ] V = π [ 128 7 − 384 21 + 384 35 − 32 + 160 3 − 8 + 8 ] V = \pi \left[ \frac{128}{7} - \frac{384}{21} + \frac{384}{35} - 32 + \frac{160}{3} - 8 + 8 \right] V = π [ 7 128 − 21 384 + 35 384 − 32 + 3 160 − 8 + 8 ] V = π [ 128 7 − 1808 105 + 13728 245 − 32 + 160 3 ] V = \pi \left[ \frac{128}{7} - \frac{1808}{105} + \frac{13728}{245} - 32 + \frac{160}{3} \right] V = π [ 7 128 − 105 1808 + 245 13728 − 32 + 3 160 ] V = π [ 183040 1715 ] V = \pi \left[ \frac{183040}{1715} \right] V = π [ 1715 183040 ] Hence, the exact volume of the solid generated when the region R is revolved about the x-axis is [ \frac{183040}{1715}\pi ].