Math  /  Calculus

QuestionQuestion 20 0/1 pt 3 98 Details
Set up the definite integral required to find the area of the region between the graph of y=x211y=x^{2}-11 and y=2x+4y=-2 x+4. \square \int \square dxd x \square Question Help: Video Written Example Submit Question Jump to Answer
Question 21 0/10 / 1 pt 3983 \rightleftarrows 98 Details
Set up the definite integral required to find the area of the region between the graph of y=13x2y=13-x^{2} and y=5x37y=5 x-37 over the interval 7x4-7 \leq x \leq 4. \square \square dx\int \square d x Question Help: Video Written Example Submit Question Jump to Answer

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between two curvy lines, which we can do by integrating the difference between the higher and lower lines over the right interval! Watch out! Make sure we figure out which function is on top and which is below, and also where the lines intersect to define our integration bounds!

STEP 2

1. Find Intersection Points
2. Determine Top and Bottom Functions
3. Set Up the Integral
4. Compute the Definite Integral (Question 20 only)

STEP 3

Let's set the two equations equal to each other: y=x211y = x^2 - 11 and y=2x+4y = -2x + 4.
So, x211=2x+4x^2 - 11 = -2x + 4.

STEP 4

Moving everything to one side, we get x2+2x15=0x^2 + 2x - 15 = 0.
Factoring this **quadratic equation** gives us (x+5)(x3)=0(x+5)(x-3) = 0.

STEP 5

This means our **intersection points** are at x=5x = \mathbf{-5} and x=3x = \mathbf{3}!

STEP 6

The problem already gives us the interval: 7x4-7 \leq x \leq 4.
So, our **bounds of integration** will be from 7\mathbf{-7} to 4\mathbf{4}!

STEP 7

Let's test a point between our intersection points, like x=0x = 0.
Plugging into y=x211y = x^2 - 11, we get y=11y = -11.
Plugging into y=2x+4y = -2x + 4, we get y=4y = 4.

STEP 8

Since 11<4-11 < 4, the line y=2x+4y = -2x + 4 is **above** the parabola y=x211y = x^2 - 11 in our interval.

STEP 9

Let's test x=0x = 0 again.
For y=13x2y = 13 - x^2, we get y=13y = 13.
For y=5x37y = 5x - 37, we get y=37y = -37.

STEP 10

Since 13>3713 > -37, the parabola y=13x2y = 13 - x^2 is **above** the line y=5x37y = 5x - 37 in our interval.

STEP 11

The **integral** for the area is the integral from 5-5 to 33 of the **top function** minus the **bottom function**: 53((2x+4)(x211))dx\int_{-5}^{3} ((-2x + 4) - (x^2 - 11)) \, dx.

STEP 12

Simplifying, we get 53(x22x+15)dx\int_{-5}^{3} (-x^2 - 2x + 15) \, dx.

STEP 13

The **integral** for this area is 74((13x2)(5x37))dx\int_{-7}^{4} ((13 - x^2) - (5x - 37)) \, dx.

STEP 14

Simplifying, we get 74(x25x+50)dx\int_{-7}^{4} (-x^2 - 5x + 50) \, dx.

STEP 15

Evaluating the integral, we have 53(x22x+15)dx=[13x3x2+15x]53\int_{-5}^{3} (-x^2 - 2x + 15) \, dx = \left[ -\frac{1}{3}x^3 - x^2 + 15x \right]_{-5}^{3}

STEP 16

Plugging in the bounds, we get (13(3)3(3)2+15(3))(13(5)3(5)2+15(5))\left( -\frac{1}{3}(\mathbf{3})^3 - (\mathbf{3})^2 + 15(\mathbf{3}) \right) - \left( -\frac{1}{3}(\mathbf{-5})^3 - (\mathbf{-5})^2 + 15(\mathbf{-5}) \right)

STEP 17

=(99+45)(12532575)=27(1753)=27+1753=81+1753=2563= (-9 - 9 + 45) - (\frac{125}{3} - 25 - 75) = 27 - (-\frac{175}{3}) = 27 + \frac{175}{3} = \frac{81 + 175}{3} = \frac{\mathbf{256}}{\mathbf{3}}.

STEP 18

Question 20: 53(x22x+15)dx=2563\int_{-5}^{3} (-x^2 - 2x + 15) \, dx = \frac{256}{3} Question 21: 74(x25x+50)dx\int_{-7}^{4} (-x^2 - 5x + 50) \, dx

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