Find the value of \int_4^5 \,logx \,dx.
(a) 5 log5-log4+1
(b) 5 log5-4 log4-1
(c) 4 log5-4 log4-1
(d) 5-4 log4-log5
I got this question by my college director while I was bunking the class.
This question is from Fundamental Theorem of Calculus-1 topic in division Integrals of Mathematics – Class 12
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The correct answer is (b) 5 log5-4 log4-1
For explanation I would say: Let I=\int_4^5 \,logx \,dx.
F(x)=∫ logx dx
By using the formula \int \,u.v dx=u \int v \,dx-\int u'(\int \,v \,dx), we get
\int log x \,dx=logx \int \,dx-\int(logx)’\int \,dx
F(x)=x logx-∫ dx=x(logx-1).
Applying the limits using the fundamental theorem of calculus, we get
I=F(5)-F(4)=(5 log5-5)-(4 log4-4)
=5 log5-4 log4-1.