AP EAMCET · Maths · Definite Integration
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 \(\int_0^2 f(x) d x=\)
- A \(\frac{1}{2} \log 3+\frac{10}{3}\)
- B \(\frac{1}{2} \log 3-\frac{10}{3}\)
- C \(\frac{1}{2} \log 3+\frac{13}{3}\)
- D \(\frac{1}{2} \log 3+\frac{20}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \log 3+\frac{10}{3}\)
Step-by-step Solution
Detailed explanation
\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…
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