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Worked Examples · Example 14

Q.Evaluate ∫−11exex+e−x dx\int_{-1}^{1} \frac{e^x}{e^x+e^{-x}}\,dx

Sikkim CbseNCERTSubjective· 3mImportance★★★★★
24% · 14/59 Questions
✓ Free question

Use the symmetric-interval property; the integrand plus its reflection sums to 11, giving I=1I=1.

∫−aaf(x) dx=∫−aaf(−x) dx\displaystyle\int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx. If f(x)+f(−x)=kf(x)+f(-x)=k (constant), then 2I=∫−aak dx2I=\int_{-a}^{a}k\,dx.

  1. Let I=∫−11exex+e−x dxI=\displaystyle\int_{-1}^{1}\frac{e^x}{e^x+e^{-x}}\,dx and f(x)=exex+e−xf(x)=\dfrac{e^x}{e^x+e^{-x}}.
  2. Replace x→−xx\to -x: f(−x)=e−xe−x+exf(-x)=\dfrac{e^{-x}}{e^{-x}+e^{x}}.
  3. Add: f(x)+f(−x)=ex+e−xex+e−x=1.f(x)+f(-x)=\dfrac{e^x+e^{-x}}{e^x+e^{-x}}=1.
  4. Hence 2I=∫−11(f(x)+f(−x)) dx=∫−111 dx=[x]−11=2.2I=\displaystyle\int_{-1}^{1}\big(f(x)+f(-x)\big)\,dx=\int_{-1}^{1}1\,dx=\big[x\big]_{-1}^{1}=2.
  5. ∴I=22=1.\therefore I=\dfrac{2}{2}=1.
✓Final answer

∫−11exex+e−x dx=1.\displaystyle\int_{-1}^{1}\frac{e^x}{e^x+e^{-x}}\,dx=1.

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