Calculate The Area Bounded By The Graphs

\begin{aligned} 1. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 4 and g (x) = x + 2 on the interval [- 2,4] . \end{aligned}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} - 4 = x + 2 \end{aligned}

Setting the function equal to zero:

\begin{aligned} x^{2} - x - 6 = 0 \end{aligned}

Factoring, we get:

\begin{array}{r} (x + 2) (x - 3) = 0 \end{array}

Solving, we get:

\begin{array}{r} x = - 2 a n d x = 3 \end{array}

The question is asking to find the area in the interval [-2,4]. So we will find the area from x=-2 to x=3 and add it to the area from x=3 to x=4

\begin{aligned} Area = \int_{- 2}^{3} [g (x) - f (x)] d x + \int_{3}^{4} [f (x) - g (x)] d x \\ = \int_{- 2}^{3} [(x + 2) - (x^{2} - 4)] d x + \int_{3}^{4} [(x^{2} - 4) - (x + 2)] d x \\ = \int_{- 2}^{3} (x + 2 - x^{2} + 4) d x + \int_{3}^{4} (x^{2} - 4 - x - 2) d x \\ = \int_{- 2}^{3} (- x^{2} + x + 6) d x + \int_{3}^{4} (x^{2} - x - 6) d x \\ = {[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x]}_{- 2}^{3} + {[\frac{x^{3}}{3} - \frac{x^{2}}{2} - 6 x]}_{3}^{4} \\ = [(- \frac{(3)^{3}}{3} + \frac{(3)^{2}}{2} + 6 (3)) - (- \frac{(- 2)^{3}}{3} + \frac{(- 2)^{2}}{2} + 6 (- 2))] + [(\frac{(4)^{3}}{3} - \frac{(4)^{2}}{2} - 6 (4)) - (\frac{(3)^{3}}{3} - \frac{(3)^{2}}{2} - 6 (3))] \\ = [(- \frac{27}{3} + \frac{9}{2} + 18) - (- \frac{- 8}{3} + \frac{4}{2} - 12))] + [(\frac{64}{3} - \frac{16}{2} - 24) - (\frac{27}{3} - \frac{9}{2} - 18)] \\ = [(\frac{27}{2}) - (- \frac{22}{3})] + [(- \frac{32}{3}) - (- \frac{27}{2})] \\ = (\frac{125}{6}) + (\frac{17}{6}) \\ = \frac{142}{6} \\ = \frac{71}{3} \end{aligned}

Concordia – April 2024 Final

\begin{aligned} 2. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 81 and g (x) = - 2 x - 1 on the interval - 1 \leq x \leq 10 . \end{aligned}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{array}{r} f (x) = g (x) \\ x^{2} - 81 = - 2 x - 1 \end{array}

Setting the function equal to zero:

\begin{array}{r} x^{2} + 2 x - 80 = 0 \end{array}

Factoring, we get:

\begin{array}{r} (x + 10) (x - 8) = 0 \end{array}

Solving, we get:

\begin{array}{r} x = - 10 and x = 8 \end{array}

The question is asking to find the area in the interval [−1,10]. So we will find the area from x = −1 to x = 8 and add it to the area from x = 8 to x = 10

\begin{aligned} Area = \int_{- 1}^{8} [g (x) - f (x)] d x + \int_{8}^{10} [f (x) - g (x)] d x \\ = \int_{- 1}^{8} [(- 2 x - 1) - (x^{2} - 81)] d x + \int_{8}^{10} [(x^{2} - 81) - (- 2 x - 1)] d x \\ = \int_{- 1}^{8} (- 2 x - 1 - x^{2} + 81) d x + \int_{8}^{10} (x^{2} - 81 + 2 x + 1) d x \\ = \int_{- 1}^{8} (- x^{2} - 2 x + 80) d x + \int_{8}^{10} (x^{2} + 2 x - 80) d x \\ = {[- \frac{x^{3}}{3} - \frac{2 x^{2}}{2} + 80 x]}_{- 1}^{8} + {[\frac{x^{3}}{3} + \frac{2 x^{2}}{2} - 80 x]}_{8}^{10} \\ = {[- \frac{x^{3}}{3} - x^{2} + 80 x]}_{- 1}^{8} + {[\frac{x^{3}}{3} + x^{2} - 80 x]}_{8}^{10} \\ = [(- \frac{(8)^{3}}{3} - (8)^{2} + 80 (8)) - (- \frac{(- 1)^{3}}{3} - (- 1)^{2} + 80 (- 1))] + [(\frac{(10)^{3}}{3} + (10)^{2} - 80 (10)) - (\frac{(8)^{3}}{3} + (8)^{2} - 80 (8))] \\ = [(- \frac{512}{3} - 64 + 640) - (- \frac{- 1}{3} - 1 - 80))] + [(\frac{1000}{3} + 100 - 800) - (\frac{512}{3} - 64 - 640)] \\ = [(\frac{1216}{3}) - (- \frac{242}{3})] + [(- \frac{1100}{3}) - (- \frac{1216}{3})] \\ = (\frac{1458}{3}) + (\frac{116}{3}) \\ = 486 + \frac{116}{3} \\ = \frac{1574}{3} \end{aligned}

Concordia – April 2019 Final

\begin{array}{l} 3. Calculate the area bounded by the graphs of the following two functions \\ f (x) = 2 x^{2} and g (x) = 4 - 2 x on the interval [−2,2] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{array}{r} f (x) = g (x) \\ 2 x^{2} = 4 - 2 x \end{array}

Setting the function equal to zero:

\begin{array}{r} 2 x^{2} + 2 x - 4 = 0 \end{array}

Divide by 2:

\begin{array}{r} x^{2} + x - 2 = 0 \end{array}

Factoring, we get:

\begin{array}{r} (x + 2) (x - 1) = 0 \end{array}

Solving, we get:

\begin{array}{r} x = - 2 a n d x = 1 \end{array}

The question is asking to find the area in the interval [−2,2]. So we will find the area from x = −2 to x = 1 and add it to the area from x = 1 to x = 2

\begin{aligned} Area = \int_{- 2}^{1} [g (x) - f (x)] d x + \int_{1}^{2} [f (x) - g (x)] d x \\ = \int_{- 2}^{1} [(4 - 2 x) - (2 x^{2})] d x + \int_{1}^{2} [(2 x^{2}) - (4 - 2 x)] d x \\ = \int_{- 2}^{1} (4 - 2 x - 2 x^{2}) d x + \int_{1}^{2} (2 x^{2} - 4 + 2 x) d x \\ = \int_{- 2}^{1} (- 2 x^{2} - 2 x + 4) d x + \int_{1}^{2} (2 x^{2} + 2 x - 4) d x \\ = {[- \frac{2 x^{3}}{3} - \frac{2 x^{2}}{2} + 4 x]}_{- 2}^{1} + {[\frac{2 x^{3}}{3} + \frac{2 x^{2}}{2} - 4 x]}_{1}^{2} \\ = {[- \frac{2 x^{3}}{3} - x^{2} + 4 x]}_{- 2}^{1} + {[\frac{2 x^{3}}{3} + x^{2} - 4 x]}_{1}^{2} \\ = [(- \frac{2 (1)^{3}}{3} - (1)^{2} + 4 (1)) - (- \frac{2 (- 2)^{3}}{3} - (- 2)^{2} + 4 (- 2))] + [(\frac{2 (2)^{3}}{3} + (2)^{2} - 4 (2)) - (\frac{2 (1)^{3}}{3} + (1)^{2} - 4 (1))] \\ = [(- \frac{2}{3} - 1 + 4) - (- \frac{- 16}{3} - 4 - 8)] + [(\frac{16}{3} + 4 - 8) - (\frac{2}{3} + 1 - 4)] \\ = [(\frac{7}{3}) - (- \frac{20}{3})] + [(\frac{4}{3}) - (- \frac{7}{3})] \\ = (\frac{27}{3}) + (\frac{11}{3}) \\ = \frac{38}{3} \end{aligned}

\begin{array}{l} 4. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 25 and g (x) = - 3 x + 3 on the interval [−2,6] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} - 25 = - 3 x + 3 \end{aligned}

Setting the function equal to zero:

x^{2} + 3 x - 28 = 0

Factoring, we get:

(x + 7) (x - 4) = 0

Solving, we get:

x = - 7 a n d x = 4

The question is asking to find the area in the interval [−2,6]. So we will find the area from x = −2 to x = 4 and add it to the area from x = 4 to x = 6

\begin{aligned} Area = \int_{- 2}^{4} [g (x) - f (x)] d x + \int_{4}^{6} [f (x) - g (x)] d x \\ = \int_{- 2}^{4} [(- 3 x + 3) - (x^{2} - 25)] d x + \int_{4}^{6} [(x^{2} - 25) - (- 3 x + 3)] d x \\ = \int_{- 2}^{4} (- 3 x + 3 - x^{2} + 25) d x + \int_{4}^{6} (x^{2} - 25 + 3 x - 3) d x \\ = \int_{- 2}^{4} (- x^{2} - 3 x + 28) d x + \int_{4}^{6} (x^{2} + 3 x - 28) d x \\ = {[- \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 28 x]}_{- 2}^{4} + {[\frac{x^{3}}{3} + \frac{3 x^{2}}{2} - 28 x]}_{4}^{6} \\ = [(- \frac{(4)^{3}}{3} - \frac{3 (4)^{2}}{2} + 28 (4)) - (- \frac{(- 2)^{3}}{3} - \frac{3 (- 2)^{2}}{2} + 28 (- 2))] + [(\frac{(6)^{3}}{3} + \frac{3 (6)^{2}}{2} - 28 (6)) - (\frac{(4)^{3}}{3} + \frac{3 (4)^{2}}{2} - 28 (4))] \\ = [(- \frac{64}{3} - \frac{48}{2} + 112) - (- \frac{- 8}{3} - \frac{12}{2} - 56))] + [(\frac{216}{3} + \frac{108}{2} - 168) - (\frac{64}{3} + \frac{48}{2} - 112)] \\ = [(\frac{200}{3}) - (- \frac{178}{3})] + [(- 42) - (- \frac{200}{3})] \\ = (126) + (\frac{74}{3}) \\ = \frac{452}{3} \end{aligned}

\begin{array}{l} 5. Calculate the area bounded by the graphs of the following two functions \\ f (x) = 2 x^{2} and g (x) = - 4 x + 16 on the interval [−2,4] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ 2 x^{2} = - 4 x + 16 \end{aligned}

Setting the function equal to zero:

2 x^{2} + 4 x - 16 = 0

Divide by 2:

x^{2} + 2 x - 8 = 0

Factoring, we get:

(x + 4) (x - 2) = 0

Solving, we get:

x = - 4 a n d x = 2

The question is asking to find the area in the interval [−2,4]. So we will find the area from x = −2 to x = 2 and add it to the area from x = 2 to x = 4

\begin{array}{r} Area = \int_{- 2}^{2} [g (x) - f (x)] d x + \int_{2}^{4} [f (x) - g (x)] d x \\ = \int_{- 2}^{2} [(- 4 x + 16) - (2 x^{2})] d x + \int_{2}^{4} [(2 x^{2}) - (- 4 x + 16)] d x \\ = \int_{- 2}^{2} (- 4 x + 16 - 2 x^{2}) d x + \int_{2}^{4} (2 x^{2} + 4 x - 16) d x \\ = \int_{- 2}^{2} (- 2 x^{2} - 4 x + 16) d x + \int_{2}^{4} (2 x^{2} + 4 x - 16) d x \\ = {[- \frac{2 x^{3}}{3} - \frac{4 x^{2}}{2} + 16 x]}_{- 2}^{2} + {[\frac{2 x^{3}}{3} + \frac{4 x^{2}}{2} - 16 x]}_{2}^{4} \\ = {[- \frac{2 x^{3}}{3} - 2 x^{2} + 16 x]}_{- 2}^{2} + {[\frac{2 x^{3}}{3} + 2 x^{2} - 16 x]}_{2}^{4} \\ = [(- \frac{2 (2)^{3}}{3} - 2 (2)^{2} + 16 (2)) - (- \frac{2 (- 2)^{3}}{3} - 2 (- 2)^{2} + 16 (- 2))] + [(\frac{2 (4)^{3}}{3} + 2 (4)^{2} - 16 (4)) - (\frac{2 (2)^{3}}{3} + 2 (2)^{2} - 16 (2))] \\ = [(- \frac{16}{3} - 8 + 32) - (- \frac{- 16}{3} - 8 - 32)] + [(\frac{128}{3} + 32 - 64) - (\frac{16}{3} + 8 - 32)] \\ = [(\frac{56}{3}) - (- \frac{104}{3})] + [(\frac{32}{3}) - (- \frac{56}{3})] \\ = (\frac{160}{3}) + (\frac{88}{3}) \\ = \frac{248}{3} \end{array}

\begin{array}{l} 6. Calculate the area bounded by the graphs of the following two functions \\ f (x) = 2 x^{2} - 10 and g (x) = - 4 x + 20 on the interval [−1,5] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ 2 x^{2} - 10 = - 4 x + 20 \end{aligned}

Setting the function equal to zero:

2 x^{2} + 4 x - 30 = 0

Divide by 2:

x^{2} + 2 x - 15 = 0

Factoring, we get:

(x + 5) (x - 3) = 0

Solving, we get:

x = - 5 a n d x = 3

The question is asking to find the area in the interval [−1,5]. So we will find the area from x = −1 to x = 3 and add it to the area from x = 3 to x = 5

\begin{aligned} Area = \int_{- 1}^{3} [g (x) - f (x)] d x + \int_{3}^{5} [f (x) - g (x)] d x \\ = \int_{- 1}^{3} [(- 4 x + 20) - (2 x^{2} - 10)] d x + \int_{3}^{5} [(2 x^{2} - 10) - (- 4 x + 20)] d x \\ = \int_{- 1}^{3} (- 4 x + 20 - 2 x^{2} + 10) d x + \int_{3}^{5} (2 x^{2} - 10 + 4 x - 20) d x \\ = \int_{- 1}^{3} (- 2 x^{2} - 4 x + 30) d x + \int_{3}^{5} (2 x^{2} + 4 x - 30) d x \\ = {[- \frac{2 x^{3}}{3} - \frac{4 x^{2}}{2} + 30 x]}_{- 1}^{3} + {[\frac{2 x^{3}}{3} + \frac{4 x^{2}}{2} - 30 x]}_{3}^{5} \\ = {[- \frac{2 x^{3}}{3} - 2 x^{2} + 30 x]}_{- 1}^{3} + {[\frac{2 x^{3}}{3} + 2 x^{2} - 30 x]}_{3}^{5} \\ = [(- \frac{2 (3)^{3}}{3} - 2 (3)^{2} + 30 (3)) - (- \frac{2 (- 1)^{3}}{3} - 2 (- 1)^{2} + 30 (- 1))] + [(\frac{2 (5)^{3}}{3} + 2 (5)^{2} - 30 (5)) - (\frac{2 (3)^{3}}{3} + 2 (3)^{2} - 30 (3))] \\ = [(- \frac{54}{3} - 18 + 90) - (- \frac{- 2}{3} - 2 - 30)] + [(\frac{250}{3} + 50 - 150) - (\frac{54}{3} + 18 - 90)] \\ = [(54) - (- \frac{94}{3})] + [(- \frac{50}{3}) - (- 54)] \\ = (\frac{256}{3}) + (\frac{112}{3}) \\ = \frac{368}{3} \end{aligned}

\begin{array}{l} 7. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 14 and g (x) = - 5 x - 8 on the interval [−3,4] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} - 14 = - 5 x - 8 \end{aligned}

Setting the function equal to zero:

x^{2} + 5 x - 6 = 0

Factoring, we get:

(x + 6) (x - 1) = 0

Solving, we get:

x = - 6 a n d x = 1

The question is asking to find the area in the interval [−3,4]. So we will find the area from x = −3 to x = 1 and add it to the area from x = 1 to x = 4

\begin{aligned} Area = \int_{- 3}^{1} [g (x) - f (x)] d x + \int_{1}^{4} [f (x) - g (x)] d x \\ = \int_{- 3}^{1} [(- 3 x + 3) - (x^{2} - 25)] d x + \int_{1}^{4} [(x^{2} - 25) - (- 3 x + 3)] d x \\ = \int_{- 3}^{1} (- 3 x + 3 - x^{2} + 25) d x + \int_{1}^{4} (x^{2} - 25 + 3 x - 3) d x \\ = \int_{- 3}^{1} (- x^{2} - 3 x + 28) d x + \int_{1}^{4} (x^{2} + 3 x - 28) d x \\ = {[- \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 28 x]}_{- 3}^{1} + {[\frac{x^{3}}{3} + \frac{3 x^{2}}{2} - 28 x]}_{1}^{4} \\ : [(- \frac{(4)^{3}}{3} - \frac{3 (4)^{2}}{2} + 28 (4)) - (- \frac{(- 2)^{3}}{3} - \frac{3 (- 2)^{2}}{2} + 28 (- 2))] + [(\frac{(6)^{3}}{3} + \frac{3 (6)^{2}}{2} - 28 (6)) - (\frac{(4)^{3}}{3} + \frac{3 (4)^{2}}{2} - 28 (4))] \\ = [(- \frac{64}{3} - \frac{48}{2} + 112) - (- \frac{- 8}{3} - \frac{12}{2} - 56))] + [(\frac{216}{3} + \frac{108}{2} - 168) - (\frac{64}{3} + \frac{48}{2} - 112)] \\ = [(\frac{200}{3}) - (- \frac{178}{3})] + [(- 42) - (- \frac{200}{3})] \\ = (126) + (\frac{74}{3}) \\ = \frac{452}{3} \end{aligned}

\begin{array}{l} 8. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} + 2 x - 3 and g (x) = 2 x + 6 on the interval [−1,4] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} + 2 x - 3 = 2 x + 6 \end{aligned}

Setting the function equal to zero:

x^{2} - 9 = 0

Factoring, we get:

(x + 3) (x - 3) = 0

Solving, we get:

x = - 3 a n d x = 3

The question is asking to find the area in the interval [−1,4]. So we will find the area from x = −1 to x = 3 and add it to the area from x = 3 to x = 4

\begin{aligned} Area = \int_{- 1}^{3} [g (x) - f (x)] d x + \int_{3}^{4} [f (x) - g (x)] d x \\ = \int_{- 1}^{3} [(2 x + 6) - (x^{2} + 2 x - 3)] d x + \int_{3}^{4} [(x^{2} + 2 x - 3) - (2 x + 6)] d x \\ = \int_{- 1}^{3} (2 x + 6 - x^{2} - 2 x + 3) d x + \int_{3}^{4} (x^{2} + 2 x - 3 - 2 x - 6) d x \\ = \int_{- 1}^{3} (- x^{2} + 9) d x + \int_{3}^{4} (x^{2} - 9) d x \\ = {[- \frac{x^{3}}{3} + 9 x]}_{- 1}^{3} + {[\frac{x^{3}}{3} - 9 x]}_{3}^{4} \\ = [(- \frac{(3)^{3}}{3} + 9 (3)) - (- \frac{(- 1)^{3}}{3} + 9 (- 1))] + [(\frac{(4)^{3}}{3} - 9 (4)) - (\frac{(3)^{3}}{3} - 9 (3))] \\ = [(- \frac{27}{3} + 27) - (- \frac{- 1}{3} - 9))] + [(\frac{64}{3} - 36) - (\frac{27}{3} - 27)] \\ = [(18) - (- \frac{26}{3})] + [(- \frac{44}{3}) - (- 18)] \\ = (\frac{80}{3}) + (\frac{10}{3}) \\ = 30 \end{aligned}

\begin{array}{l} 9. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 49 and g (x) = 2 x - 1 on the interval [1,10] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} - 49 = 2 x - 1 \end{aligned}

Setting the function equal to zero:

x^{2} - 2 x - 48 = 0

Factoring, we get:

(x + 6) (x - 8) = 0

Solving, we get:

x = - 6 a n d x = 8

The question is asking to find the area in the interval [1,10]. So we will find the area from x = 1 to x = 8 and add it to the area from x = 8 to x = 10

\begin{aligned} Area = \int_{1}^{8} [g (x) - f (x)] d x + \int_{8}^{10} [f (x) - g (x)] d x \\ = \int_{1}^{8} [(2 x - 1) - (x^{2} - 49)] d x + \int_{8}^{10} [(x^{2} - 49) - (2 x - 1)] d x \\ = \int_{1}^{8} (2 x - 1 - x^{2} + 49) d x + \int_{8}^{10} (x^{2} - 49 - 2 x + 1) d x \\ = \int_{1}^{8} (- x^{2} + 2 x + 48) d x + \int_{8}^{10} (x^{2} - 2 x - 48) d x \\ = {[- \frac{x^{3}}{3} + \frac{2 x^{2}}{2} + 48 x]}_{1}^{8} + {[\frac{x^{3}}{3} - \frac{2 x^{2}}{2} - 48 x]}_{8}^{10} \\ = {[- \frac{x^{3}}{3} + x^{2} + 48 x]}_{1}^{8} + {[\frac{x^{3}}{3} - x^{2} - 48 x]}_{8}^{10} \\ = [(- \frac{(8)^{3}}{3} + (8)^{2} + 48 (8)) - (- \frac{(1)^{3}}{3} + (1)^{2} + 48 (1))] + [(\frac{(10)^{3}}{3} - (10)^{2} - 48 (10)) - (\frac{(8)^{3}}{3} - (8)^{2} - 48 (4))] \\ = [(- \frac{512}{3} + 64 + 384) - (- \frac{1}{3} + 1 + 48)] + [(\frac{1000}{3} - 100 - 480) - (\frac{512}{3} - 64 - 384)] \\ = [(\frac{832}{3}) - (\frac{146}{3})] + [(- \frac{740}{3}) - (- \frac{832}{3})] \\ = (\frac{686}{3}) + (\frac{92}{3}) \\ = \frac{778}{3} \end{aligned}

\begin{array}{l} 10. Calculate the area bounded by the graphs of the following two functions \\ f (x) = x^{2} - 3 x - 15 and g (x) = 2 x - 1 on the interval [3,9] . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ x^{2} - 3 x - 15 = 2 x - 1 \end{aligned}

Setting the function equal to zero:

x^{2} - 5 x - 14 = 0

Factoring, we get:

(x + 2) (x - 7) = 0

Solving, we get:

x = - 2 a n d x = 7

The question is asking to find the area in the interval [3,9]. So we will find the area from x = 3 to x = 7 and add it to the area from x = 7 to x = 9

\begin{aligned} Area = \int_{3}^{7} [g (x) - f (x)] d x + \int_{7}^{9} [f (x) - g (x)] d x \\ = \int_{3}^{7} [(2 x - 1) - (x^{2} - 3 x - 15)] d x + \int_{7}^{9} [(x^{2} - 3 x - 15) - (2 x - 1)] d x \\ = \int_{3}^{7} (2 x - 1 - x^{2} + 3 x + 15) d x + \int_{7}^{9} (x^{2} - 3 x - 15 - 2 x + 1) d x \\ = \int_{3}^{7} (- x^{2} + 5 x + 14) d x + \int_{7}^{9} (x^{2} - 5 x - 14) d x \\ = {[- \frac{x^{3}}{3} + \frac{5 x^{2}}{2} + 14 x]}_{3}^{7} + {[\frac{x^{3}}{3} - \frac{5 x^{2}}{2} - 14 x]}_{7}^{9} \\ = [(- \frac{(7)^{3}}{3} + \frac{5 (7)^{2}}{2} + 14 (7)) - (- \frac{(3)^{3}}{3} + \frac{5 (3)^{2}}{2} + 14 (3))] + [(\frac{(9)^{3}}{3} - \frac{5 (9)^{2}}{2} - 14 (9)) - (\frac{(7)^{3}}{3} - \frac{5 (7)^{2}}{2} - 14 (7))] \\ = [(- \frac{343}{3} + \frac{245}{2} + 98) - (- \frac{27}{3} + \frac{45}{2} + 42))] + [(\frac{729}{3} - \frac{405}{2} - 126) - (\frac{343}{3} - \frac{245}{2} - 98)] \\ = [(\frac{637}{6}) - (\frac{111}{2})] + [(- \frac{171}{2}) - (- \frac{637}{6})] \\ = (\frac{152}{3}) + (\frac{62}{3}) \\ = \frac{214}{3} \end{aligned}

\begin{array}{l} 11. Calculate the area bounded by the graphs of the following two functions \\ f (x) = 4 x - 6 x^{2} + 34 and g (x) = 4 x + 10 for - 1 \leq x \leq 3 . \end{array}

(accurate to three decimal places)

First, we equate the functions to find the points of intersection.

\begin{aligned} f (x) = g (x) \\ 4 x - 6 x^{2} + 34 = 4 x + 10 \end{aligned}

Setting the function equal to zero:

6 x^{2} - 24 = 0

Factoring, we get:

6 (x + 2) (x - 2) = 0

Solving, we get:

x = - 2 a n d x = 2

The question is asking to find the area in the interval [−1,3]. So we will find the area from x = −1 to x = 2 and add it to the area from x = 2 to x = 3

\begin{aligned} Area 1 = \int_{- 1}^{2} [f (x) - g (x)] d x + \int_{2}^{3} [g (x) - f (x)] d x \\ = \int_{- 1}^{2} [(4 x - 6 x^{2} + 34) - (4 x + 10)] d x + \int_{2}^{3} [(4 x + 10) - (4 x - 6 x^{2} + 34)] d x \\ = \int_{- 1}^{2} (4 x - 6 x^{2} + 34 - 4 x - 10) d x + \int_{2}^{3} (4 x + 10 - 4 x + 6 x^{2} - 34) d x \\ = \int_{- 1}^{2} (- 6 x^{2} + 24) d x + \int_{2}^{3} (6 x^{2} - 24) d x \\ = {[- \frac{6 x^{3}}{3} + 24 x]}_{- 1}^{2} + {[\frac{6 x^{3}}{3} - 24 x]}_{2}^{3} \\ = {[- \frac{2 x^{3}}{1} + 24 x]}_{- 1}^{2} + {[\frac{2 x^{3}}{1} - 24 x]}_{2}^{3} \\ [(- 2 (2)^{3} + 24 (2)) - (- 2 (- 1)^{3} + 24 (- 1))] + [(2 (3)^{3} - 24 (3)) - (2 (2)^{3} - 24 (2)) \\ = [(- 16 + 48) - (2 - 24)] + [(54 - 72) - (16 - 48)] \\ = [(32) - (- 22)] + [(- 18) - (- 32)] \\ \begin{array}{ccc} = & (54) & + & (14) \end{array} \\ = 68 \end{aligned}

Calculating the Area Bounded by Graphs

To calculate the area bounded by graphs, we generally use integration, especially if the graphs are represented by functions. The process involves finding the definite integral of the difference between the functions that describe the upper and lower boundaries of the area. Here’s a step-by-step explanation:

Table of Contents

Step 1: Identify the Functions and Limits

Suppose you have two functions $f(x)$ and $g(x)$ , where $f(x)$ represents the upper curve and $g(x)$ represents the lower curve. The area between these two curves from $x = a$ to $x = b$ can be found by integrating the difference $f(x) – g(x)$ over the interval $[a, b]$ .

Step 2: Set Up the Integral

The area $A$ bounded by the curves $f(x)$ and $g(x)$ from $x = a$ to $x = b$ is given by:

$A = \int_{a}^{b} [f (x) - g (x)] d x$

Step 3: Calculate the Integral

Evaluate the definite integral to find the area.

Example Problem

Let’s say you want to find the area bounded by the curves $y = x^2$ and $y = x + 2$ between their points of intersection.

Step 1: Find the Points of Intersection

Set the two functions equal to find the points where they intersect:

$x^{2} = x + 2$

Rearrange the equation:

$x^{2} - x - 2 = 0$

Factor the quadratic equation:

$(x - 2) (x + 1) = 0$

So, $x = 2$ and $x = -1$ are the points of intersection.

Step 2: Set Up the Integral

The area $A$ is:

$A = \int_{- 1}^{2} [(x + 2) - x^{2}] d x$

Step 3: Evaluate the Integral

Calculate the integral:

$A = \int_{- 1}^{2} (x + 2 - x^{2}) d x$

First, integrate each term:

$A = {[\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3}]}_{- 1}^{2}$

Now, plug in the limits $x = 2$ and $x = -1$ :

For $x = 2$ :

$\frac{2^2}{2} + 2(2) – \frac{2^3}{3} = 2 + 4 – \frac{8}{3} = 6 – \frac{8}{3} = \frac{18}{3} – \frac{8}{3} = \frac{10}{3}$

For $x = -1$ :

$\frac{(-1)^2}{2} + 2(-1) – \frac{(-1)^3}{3} = \frac{1}{2} – 2 + \frac{1}{3} = \frac{3}{6} – \frac{12}{6} + \frac{2}{6} = \frac{-7}{6}$

So the total area is:

$A = \frac{10}{3} – \left(\frac{-7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = 4.5 \text{ square units}$

Final Answer

The area bounded by the curves $y = x^2$ and $y = x + 2$ between their points of intersection is $4.5$ square units.

Calculate the area bounded by the graphs

Step 1: Identify the Functions and Limits

Step 2: Set Up the Integral

Step 3: Calculate the Integral

Example Problem

Step 1: Find the Points of Intersection

Step 2: Set Up the Integral

Step 3: Evaluate the Integral

Final Answer

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Step 1: Identify the Functions and Limits

Step 2: Set Up the Integral

Step 3: Calculate the Integral

Example Problem

Step 1: Find the Points of Intersection

Step 2: Set Up the Integral

Step 3: Evaluate the Integral

Final Answer

Leave a comment Cancel reply

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