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The graph of g consists of two straight lines and a semicircle

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The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) integral^10_0 g (x) dx 100 (b) integral^30_10 g (x) dx 100 - 50 pi (c) integral^35_0 g (x) dx 225/2 - 50 pi

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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).

we have to evaluate,

$\small \int _0^{10} g(x) \ dx$

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 = 1/2 * base * height

hence we can say that the area of right angled triangle with base 10 and height 20 is given by

A = 1/2 * 10 * 20 = 100

Hence we can say that,

$\small \int _0^{10} g(x) \ dx = area \ of \ right \ angled \ triangle \ with \ base \ 10 \ and \ height \ 20$

$\small \Rightarrow \int _0^{10} g(x) \ dx = 100$

we have to evaluate,

$\small \int _{10}^{30} g(x) \ dx$

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,

$\small A = \frac{1}{2}\pi r^2$

Hence we can say that area of a semicircle with radius 10 is given by,

$\small A = \frac{1}{2}\pi (10)^2 = \frac{1}{2}\pi * 100 = 50\pi$

But we can see that semicircle is below x axis hence we can say that,

$\small \int _{10}^{30} g(x) \ dx = - (area \ of \ the \ semicircle \ with \ radius \ 10)$

$\small \Rightarrow \int _{10}^{30} g(x) \ dx = - 50\pi$

we have to evaluate,

$\small \int _{0}^{35} g(x) \ dx$

we can say that,

$\small \int _{0}^{35} g(x) \ dx = \int _{0}^{10} g(x) \ dx + \int _{10}^{30} g(x) \ dx + \int _{30}^{35} g(x) \ dx$

we have,
$\small \small \int _0^{10} g(x) \ dx = 100, \ \small \int _{10}^{30} g(x) \ dx = - 50\pi$

Hence we can say that,

$\small \int _{0}^{35} g(x) \ dx =100 - 50\pi+ \int _{30}^{35} g(x) \ dx$ ----------------------------------------------------1)

now we will evaluate,

$\small \int _{30}^{35} g(x) \ dx$

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 = 1/2 * base * height

hence we can say that the area of right angle triangle with base 5 and height 5 is given by,

A = 1/2 * 5 * 5 = 25/2

Hence we can say that,

$\small \int _{30}^{35} g(x) \ dx = area \ of \ right \ angled \ triangle \ with \ base \ 5 \ and \ height \ 5$

$\small \Rightarrow \int _{30}^{35} g(x) \ dx = \frac{25}{2}$

Put this value in equation 1) we can say that,

$\small \int _{0}^{35} g(x) \ dx =100 - 50\pi+ \frac{25}{2}$

$\small \Rightarrow \int _{0}^{35} g(x) \ dx = \frac{200+25}{2}- 50\pi$

$\small \Rightarrow \int _{0}^{35} g(x) \ dx = \frac{225}{2}- 50\pi$

The graph of g consists of two straight lines and a semicircle

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