Mathematics: Find the area between y=(x-2)ex and the x-axis from x=2 to x=4.

Find the area between y=(x-2)ex and the x-axis

ivilkas23

15.01.2022

e^4+e^2

Step-by-step explanation:

It can be found by integral.

First let's find the intersection points.

(x-2)e^x = 0

Only intersection point is x = 2.

And it is asked to find the area in the interval (2, 4).

$\int\limits^4_2 (x-2)e^x \,dx=?$

We will use integration by parts.

$x-2=u\\dx=du\\e^xdx=dv\\e^x=v$

$\int\limits^4_2 (x-2)e^x \,dx=uv-\int vdu=(x-2)e^x-\int e^xdx=(x-2)e^x-e^x=\\\\=(x-3)e^x|^4_2=(4-3)e^4-(2-3)e^2=e^4+e^2$

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salinae

16.02.2020

Part A)

Using the Pythagorean theorem you can find the height:

17^2 = 8^2 + x^2

289 = 64 +x^2

x^2 = 289 - 64

x^2 = 225

x = √225

x = 15

The height is 15 cm.

Part B)

Area of the triangular side: 1/2 x 16 x 15 = 120 square cm.

There are two triangles so 120 x 2 = 240 square cm.

One side = 20 x 17 = 340

There are 2 sides so 340 x 2 = 680 square cm.

The bottom is 16 x 20 = 320 square cm.

Total cardboard needed: 240 + 680 + 320 = 1,240 square cm.

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