Evaluate the integral by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using geometry. [0,1] |9x-3|

Question

need answer asap please!

Answer 1

Draw the graph. You will see that the region consists of two right triangles. The one on the left has base 1/3 and height 3. The one on the right has base 2/3 and height 6.The region lies entirely above the x-axis. So, the value of the integral is just the sum of the areas of the two right triangles.

So, ∫_{(0 to 1)} l 9x - 3 l dx = (1/2)(1/3)(3) + (1/2)(2/3)(6) = 1/2 + 2 = 2.5

Answer 2

Hi Iqra,

Here is a graph of the function f(x) = |9x-3| on the interval [0,1]. You can draw this graph yourself by plugging in some key points, like x=0 and x=1, and you can find where it touches the x-axis by solving the equation 0 = |9x-3| (you should get x=1/3). Keep in mind that everything will be above the x-axis, since the absolute value makes everything positive.

See the graph here: https://www.desmos.com/calculator/wekamiajma

We can see that the area under the curve consists of two triangles, and we can find the area of a triangle with the formula A=bh/2.

The left triangle has a base of 1/3 and a height of 3, so its area is 0.5.

The right triangle has a base of 2/3 and a height of 6, so its area is 2.

So the total area is 0.5 + 2 = 2.5.

Hope this helps,

The Mathematic Fanatic