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Math  /  Calculus

Question(2) (3 pts) Estimate the area under the graph of f(x)=6−4x2f(x)=6-4 x^{2} from x=0x=0 to x=5x=5 using four approximating rectangles and left endpoints.

Studdy Solution

STEP 1

What is this asking? We need to approximate the area under a curve using rectangles, like building a wall with LEGO bricks!
Specifically, we're using left endpoints, so each rectangle's height is determined by the function's value at the left edge of the rectangle's base. Watch out! Don't forget that we're using *left* endpoints, not right endpoints or midpoints.
This makes a big difference in how we calculate the rectangle heights!
Also, make sure to use the correct width for each rectangle based on the interval we're covering.

STEP 2

1. Calculate Rectangle Width
2. Calculate Left Endpoints
3. Calculate Rectangle Heights
4. Calculate Rectangle Areas
5. Calculate Total Area

STEP 3

We're going from x=0x = 0 to x=5x = 5 and using **four** rectangles.
So, the width of each rectangle is the **total width** of the interval divided by the **number of rectangles**.

STEP 4

Rectangle Width=5−04=54=1.25 \text{Rectangle Width} = \frac{5 - 0}{4} = \frac{5}{4} = 1.25 So, each rectangle is **1.25** wide!

STEP 5

Since we're using left endpoints, the first endpoint is x1=0x_1 = 0.
To find the other endpoints, we keep adding the **rectangle width** we just calculated.

STEP 6

x1=0 x_1 = 0 x2=0+1.25=1.25 x_2 = 0 + 1.25 = 1.25 x3=1.25+1.25=2.5 x_3 = 1.25 + 1.25 = 2.5 x4=2.5+1.25=3.75 x_4 = 2.5 + 1.25 = 3.75 These are the left edges of our four rectangles!

STEP 7

Now, we plug each left endpoint into our function f(x)=6−4x2f(x) = 6 - 4x^2 to find the height of each rectangle.

STEP 8

f(x1)=f(0)=6−4⋅(0)2=6 f(x_1) = f(0) = 6 - 4 \cdot (0)^2 = 6 f(x2)=f(1.25)=6−4⋅(1.25)2=6−4⋅1.5625=6−6.25=−0.25 f(x_2) = f(1.25) = 6 - 4 \cdot (1.25)^2 = 6 - 4 \cdot 1.5625 = 6 - 6.25 = -0.25 f(x3)=f(2.5)=6−4⋅(2.5)2=6−4⋅6.25=6−25=−19 f(x_3) = f(2.5) = 6 - 4 \cdot (2.5)^2 = 6 - 4 \cdot 6.25 = 6 - 25 = -19 f(x4)=f(3.75)=6−4⋅(3.75)2=6−4⋅14.0625=6−56.25=−50.25 f(x_4) = f(3.75) = 6 - 4 \cdot (3.75)^2 = 6 - 4 \cdot 14.0625 = 6 - 56.25 = -50.25 These are the heights of our rectangles!
Notice some heights are negative, which means those rectangles are *below* the x-axis.

STEP 9

The area of each rectangle is its **width** times its **height**.
Remember, our width is **1.25**.

STEP 10

Area1=1.25⋅6=7.5 \text{Area}_1 = 1.25 \cdot 6 = 7.5 Area2=1.25⋅(−0.25)=−0.3125 \text{Area}_2 = 1.25 \cdot (-0.25) = -0.3125 Area3=1.25⋅(−19)=−23.75 \text{Area}_3 = 1.25 \cdot (-19) = -23.75 Area4=1.25⋅(−50.25)=−62.8125 \text{Area}_4 = 1.25 \cdot (-50.25) = -62.8125

STEP 11

Finally, we add up the areas of all the rectangles to get our estimate of the total area under the curve.

STEP 12

Total Area=7.5−0.3125−23.75−62.8125=−79.375 \text{Total Area} = 7.5 - 0.3125 - 23.75 - 62.8125 = -79.375

STEP 13

The estimated area under the graph of f(x)=6−4x2f(x) = 6 - 4x^2 from x=0x = 0 to x=5x = 5 using four approximating rectangles and left endpoints is **-79.375**.

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