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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Let $\mathrm{a}$ and $\mathrm{b}$ be real constants such that the function $f$ defined by $f(x)=\left\{\begin{array}{cc}x^2+3 x+a & x \leq 1 \\ b x+2, & x>1\end{array}\right.$ be differentiable on $R$. Then, the value of $\int_{-2}^2 f(x) d x$ equals
  • A
     $\frac{15}{6}$
  • B
    $\frac{19}{6}$
  • C
     $21$
  • ✓
    $17$
✓

Answer

Correct option: D.
$17$
d
$ \mathrm{f} \text { is continuous } \quad \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+3, \mathrm{x}<1 $

$ \therefore 4+\mathrm{a}=\mathrm{b}+2$

$ \text { b }, x>1 $

$ \mathrm{a}=\mathrm{b}-2 \quad \mathrm{f} \text { is differentiable } $

$ \therefore \mathrm{b}=5 $

$ \therefore \quad \mathrm{a}=3 $

$ \int_{-2}^1\left(x^2+3 x+3\right) d x+\int_1^2(5 x+2) d x $

$ =\left[\frac{\mathrm{x}^3}{3}+\frac{3 \mathrm{x}^2}{2}+3 \mathrm{x}\right]_{-2}^1+\left[\frac{5 \mathrm{x}^2}{2}+2 \mathrm{x}\right]_1^2$

$ =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)$

$ =6+\frac{3}{2}+12-\frac{5}{2}=17 $

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