JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of the region \(\left\{(x, y): 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1\right.\) \(\left.\frac{1}{2} \leq x \leq 2\right\}\) is
- A \(\frac{79}{16}\)
- B \(\frac{23}{6}\)
- C \(\frac{79}{24}\)
- D \(\frac{23}{16}\)
Answer & Solution
Correct Answer
(C) \(\frac{79}{24}\)
Step-by-step Solution
Detailed explanation
\(0 \leq y \leq x^{2}+1,0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2\) Required area \(=\int_{1 / 2}^{1}\left(x^{2}+1\right) d x+\frac{1}{2}(2+3) \times 1\) \(=\frac{19}{24}+\frac{5}{2}=\frac{79}{24}\)
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