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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int\limits_{ - 1}^0 {\frac{{4{x^2} + 4x + 3}}{{1 + {e^{2x + 1}}}}} dx\, = $
  • A
    $\frac{7}{3}$
  • B
    $0$
  • ✓
    $\frac{7}{6}$
  • D
    $\frac{7}{12}$
✓

Answer

Correct option: C.
$\frac{7}{6}$
c
$\int_{-1}^{0} \frac{(2 x+1)^{2}+2}{1+e^{2 x+1}}$

$\text { Put } 2 x+1=t, d x=\frac{d t}{2}$

$I=\frac{1}{2} \int_{-1}^{1} \frac{t^{2}+2}{1+e^{t}}$

$2 \mathrm{I}=\frac{1}{2} \int_{-1}^{1}\left(\mathrm{t}^{2}+2\right) \mathrm{dt}$

$2 \mathrm{I}=\frac{1}{2} \times 2 \int_{0}^{1}\left(\mathrm{t}^{2}+2\right) \mathrm{dt}$

$2 \mathrm{I}=\left(\frac{t^{3}}{3}+2 t\right)_{0}^{1}$

$I=\frac{7}{6}$

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