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3.4 · Q2

Q.Evaluate the following definite integral: ∫12e−log⁡x dx\int_1^2 e^{-\log x}\,dx

Puducherry CbseNCERTSubjective· 2mImportance★★★★★
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✓ Free question

e−log⁡x=1xe^{-\log x}=\frac1x, so the integral is ∫12dxx=log⁡2\int_1^2\frac{dx}{x}=\log 2.

Log–exp inverse: e−log⁡x=elog⁡(1/x)=1xe^{-\log x}=e^{\log(1/x)}=\dfrac1x; and ∫dxx=log⁡∣x∣\int\frac{dx}{x}=\log|x|.

  1. Given: ∫12e−log⁡x dx\displaystyle\int_1^2 e^{-\log x}\,dx.
  2. e−log⁡x=1xe^{-\log x}=\dfrac1x.
  3. =∫12dxx=[log⁡x]12=log⁡2−log⁡1=log⁡2=\displaystyle\int_1^2\frac{dx}{x}=\big[\log x\big]_1^2=\log 2-\log 1=\log 2.
  4. Value: log⁡2≈0.693\log 2\approx 0.693.
✓Final answer

∫12e−log⁡x dx=log⁡2≈0.693\displaystyle\int_1^2 e^{-\log x}\,dx=\log 2\approx 0.693.

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