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3.4 · Q3

Q.Evaluate the following definite integral: ∫log⁡2log⁡42x dx\int_{\log 2}^{\log 4} 2^x\,dx

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

Apply the exponential rule ∫ax dx=axlog⁡a\int a^x\,dx=\dfrac{a^x}{\log a} with a=2a=2. Over the limits log⁡2\log 2 to log⁡4\log 4 the value is 2log⁡4−2log⁡2log⁡2≈1.44\dfrac{2^{\log 4}-2^{\log 2}}{\log 2}\approx 1.44.

Exponential: ∫ax dx=axlog⁡a+C,a>0, a≠1.\displaystyle\int a^x\,dx=\frac{a^{x}}{\log a}+C,\quad a>0,\ a\neq1.

Here log⁡\log denotes the natural logarithm, the convention used throughout this chapter of the CBSE Class 12 Applied Mathematics book.

  1. With a=2a=2: ∫log⁡2log⁡42x dx=[2xlog⁡2]log⁡2log⁡4=2log⁡4−2log⁡2log⁡2.\displaystyle\int_{\log 2}^{\log 4} 2^x\,dx=\left[\frac{2^x}{\log 2}\right]_{\log 2}^{\log 4}=\frac{2^{\log 4}-2^{\log 2}}{\log 2}.
  2. Numerically, 2log⁡2=e(log⁡2)2≈1.6172^{\log 2}=e^{(\log 2)^2}\approx 1.617 and 2log⁡4=e2(log⁡2)2≈2.6142^{\log 4}=e^{2(\log 2)^2}\approx 2.614, so the value ≈2.614−1.6170.693≈1.44.\approx\dfrac{2.614-1.617}{0.693}\approx 1.44.
✓Final answer

∫log⁡2log⁡42x dx=2log⁡4−2log⁡2log⁡2≈1.44.\displaystyle\int_{\log 2}^{\log 4} 2^x\,dx=\frac{2^{\log 4}-2^{\log 2}}{\log 2}\approx 1.44.

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