If I_{1}=\int_{e}^{e^{2}}\frac{dx}{\ln x} and I_{2}=\int_{1}^{2}\frac{e^{x}}{x}\,dx, then which one of the following is correct?
- A. I_{1}-I_{2}=0 ✓
- B. I_{1}+I_{2}=0
- C. I_{1}-2I_{2}=0
- D. 2I_{1}-I_{2}=0
Correct Answer: A. I_{1}-I_{2}=0
Explanation
In I_1, use the substitution t = \ln x \implies x = e^t \implies dx = e^t dt. The limits change from x=e \to t=1 and x=e^2 \to t=2. Therefore, I_1 = \int_1^2 \frac{e^t}{t} \,dt, which is exactly identical to I_2. Thus, I_1 - I_2 = 0.
Related questions on Calculus
- Let z=[y] and y=[x]-x, where [.] is the greatest integer function. If x is <strong>NOT</strong> an integer but positive, then what i...
- If f(x)=4x+1 and g(x)=kx+2 such that fog(x)=gof(x), then what is the value of k?
- What is the <strong>MINIMUM</strong> value of the function f(x)=\log_{10}(x^{2}+2x+11)?
- What is \int(x^{x})^{2}(1+\ln x)\,dx equal to ?
- What is \int e^{x}\{1+\ln x+x\ln x\}\,dx equal to?