Interpret the following Integrals as signed areas to obtain its value
1 answer:
Let F be the primitive of f, i.e.
Then, we have
So, we have
As for the sine, we can spare the integral: the sine is an odd function, which means that it is symmetrical with respect to the origin. Any odd function integrated over a symmetrical neighborhood of zero (i.e. any interval of the form [-n,n]) returns zero: by symmetry, there is an equal amount of positive and negative area. So, we have
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<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2><h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2><h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2><h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
Answer:
The area of each section of the circle is
Step-by-step explanation:
we know that
The area of a circle is equal to
we have
substitute
To find the area of each section divide the complete area of the circle by
so