the graph of g consists of two straight lines and a semicircle. use it to evaluate each integral12110601351.jpga)\int_0^{10}g(x)dxb)\int_{10}^{30}g(x)dxc)\int_0^{35}g(x)dx

arenceabigns 2021-05-17 Answered

the graph of g consists of two straight lines and a semicircle. use it to evaluate each integral
image
a)010g(x)dx
b)1030g(x)dx
c)035g(x)dx

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Expert Answer

Demi-Leigh Barrera
Answered 2021-05-18 Author has 97 answers

we have to evaluate the integral from the given graph
we know that integration of the function f(x) is nothing but the area under the curve f(x).
a)
we have to evaluate,
010g(x)dx
In the graph of g(x) we can see that between x = 0 and x = 10 g(x) is nothing but the right angled triangle with base 10 and height 20
we know that the area of right angled triangle is given by,
A=12 base  height
hence we can say that the area of right angled triangle with base 10 and height 20 is given by
A=121020=100
Hence we can say that,
010g(x)dx= area of right angled triangle with base 10 and height 20 
010g(x)dx=100
b)we have to evaluate,
1030g(x)dx
In the graph of g(x) we can see that between x = 10 and x = 30 g(x) is nothing but the semicircle with radius 10
we know that area of a semicircle with radius r is given by,
A=12πr2
Hence we can say that area of a semicircle with radius 10 is given by,
A=12π(10)2=12π100=50π
But we can see that semicircle is below x axis hence we can say that,
1030g(x)dx=( area of the semicircle with radius 10)
1030g(x)dx=50π
c)
we have to evaluate,
035g(x)dx

we can say that,

035g(x)dx=010g(x)dx+1030g(x)dx+3035g(x)dx

we have,

010g(x)dx=100,1030g(x)dx=50π

Hence we can say that,

035g(x)dx=10050π+3035g(x)dx

now we will evaluate,

3035g(x)dx

In the graph of g(x) we can see that between x = 30 and x = 35 g(x) is nothing but the right angled triangle with base 5 and height 5
we know that the area of right angled triangle is given by,
A=12 base  height
hence we can say that the area of right angle triangle with base 5 and height 5 is given by,
A=1255=252
Hence we can say that,
3035g(x)dx= area of right angled triangle with base 5 and height 5
3035g(x)dx=252
Put this value in equation 1) we can say that,
035g(x)dx=10050π+252
035g(x)dx=200+25250π
035g(x)dx=225250π

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