Q115 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
If $f(x)=\left\{\begin{array}{cl}\frac{6 x^2+1}{4 x^3+2 x+3}, & , 0 \lt x \lt 1 \\ x^2+1 & , 1 \leq x \leq 2\end{array}\right.$ then $\int_0^2 f(x) d x=$
- Correct answerA$\frac{1}{2} \log 3+\frac{10}{3}$
- B$\frac{1}{2} \log 3-\frac{10}{3}$
- C$\frac{1}{2} \log 3+\frac{13}{3}$
- D$\frac{1}{2} \log 3+\frac{20}{3}$
Solution
$\begin{aligned} & f(x)=\left\{\begin{array}{cc}\frac{6 x^2+1}{4 x^3+2 x+3}, & 0 \lt x \lt 1 \\ x^2+1, & 1 \leq x \leq 2 \\ \int_0^2 f(x) d x=\int_0^1 f(x) d x+\int_1^2 f(x) d x\end{array}\right. \\ & =\int_0^1 \frac{6 x^2+1}{4 x^3+2 x+3} d x+\int_1^2\left(x^2+1\right) d x\end{aligned}$
$\begin{aligned} & =\frac{1}{2} \int_0^1 \frac{12 x^2+2}{4 x^3+2 x+3} d x+\int_1^2\left(x^2+1\right) d x \\ & =\left[\frac{1}{2} \log \left(4 x^3+2 x+3\right)\right]_0^1+\left[\frac{x^3}{3}+x\right]_1^2 \\ & =\frac{1}{2} \log 9-\frac{1}{2} \log 3+\frac{8}{3}+2-\frac{1}{3}-1=\frac{1}{2} \log 3+\frac{10}{3}\end{aligned}$
Hence, the correct answer is option: 'A'.