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.