Here is a Multi choice question to Evaluate the value of function by Integration. You can find similar set of questions at TutorVista Blogs.
Topic : Integration of Function.
Below example will help you to identify correct choice and get reasons for, why remaining choices are incorrect.
Question : If 0∫xf(t)dt = x Cos Πx, then the value f(4) is
a. 0
b. 1
c. -4Π + 1
d. -3
Solution :
Choice b is correct.
Integration is as follows,
0∫xf(t)dt = x Cos Πx
d/dx(0∫xf(t)dt) = d/dx (x Cos Πx)
f(x) = x (- Sin Πx). Π + (Cos Πx). 1
f(4) = 4Π (- Sin 4Π) + (Cos 4Π)
= 0 + 1 = 1
Choice a is incorrect because of the error in substituting the value of cos 4Π as 0 instead of 1.
Choice c is incorrect because of the error in substituting the value of sin 4Π as 1 instead of 0.
Choice d is incorrect because of the error in missing Π in the first term and also substituting the value of sin 4Π as 1 instead of 0.
If you have any queries do write to us and you will surely get help from calculus help
Topic : Integration of Function.
Below example will help you to identify correct choice and get reasons for, why remaining choices are incorrect.
Question : If 0∫xf(t)dt = x Cos Πx, then the value f(4) is
a. 0
b. 1
c. -4Π + 1
d. -3
Solution :
Choice b is correct.
Integration is as follows,
0∫xf(t)dt = x Cos Πx
d/dx(0∫xf(t)dt) = d/dx (x Cos Πx)
f(x) = x (- Sin Πx). Π + (Cos Πx). 1
f(4) = 4Π (- Sin 4Π) + (Cos 4Π)
= 0 + 1 = 1
Choice a is incorrect because of the error in substituting the value of cos 4Π as 0 instead of 1.
Choice c is incorrect because of the error in substituting the value of sin 4Π as 1 instead of 0.
Choice d is incorrect because of the error in missing Π in the first term and also substituting the value of sin 4Π as 1 instead of 0.
If you have any queries do write to us and you will surely get help from calculus help
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