The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.a) \int_{0}^{6}g(x)dxb) \int_{6}^{18}g(x)dxc) \int_{0}^{21}g(x)dx

Tolnaio

Tolnaio

Answered question

2021-05-13

The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.
image
a) 06g(x)dx
b) 618g(x)dx
c) 021g(x)dx

Answer & Explanation

Szeteib

Szeteib

Skilled2021-05-14Added 102 answers

Step 1
06g(x)dx
From the graph, we see that on [0, 2], the graph of g(x) is just a straight line coming down from y=12 to y=0.
The area is just the area of the triangle, with base 6 and height 12 units
12×b×h=12×6×12=36
Since the area is above the x axis, 06g(x)dx equals the area.
So, 06g(x)dx=36
Step 2
618g(x)dx
From the graph, we see that on [6, 18] the graph of g(x) is a semi circle below the x axis with radius 6 units, whose area is
12πr2=12π62=12π0.36=18π
Since it lies below the x axis, the integral would be negative.
So, 618g(x)dx=18π
Step 3
021g(x)dx
Since 021g(x)dx=06g(x)dx+618g(x)dx+1821g(x)dx
3618π+1821g(x)dx
we just have to evaluate 1821g(x)dx
From the graph, we see that on [18, 21], g(x) is a straight line and the integral would be area of the triangle with base 3 units and height 3 units.
12×3×3=92
And as this lies above the x axis, 1821g(x)dx1821g(x)dx=92
So, 021g(x)dx=3618π+1821g(x)dx=3618π+92+81218π=16.05

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