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Mathematics Question on Inverse Trigonometric Functions

Define f (x) = min{x^2 + 1, x + 1] for.xe∈Rx e\in R. Then ∫−11f(x)dx\int\limits_{-1}^1f (x)dx is

A

44871

B

16/6

C

18/6

D

44901

Answer

44871

Explanation

Solution

f\left(x\right) =min \left\\{x^{2} +1 , x + 1\right\\}
⇒f(x)=(x+1x<0 x2+1x≥0)\Rightarrow f\left(x\right)= \begin{pmatrix}x+1&x<0\\\ x^{2}+1&x\ge0\end{pmatrix}
⇒∫−11f(x)dx=∫−10(x+1)dx+∫01(x2+1)dx\Rightarrow \int_{-1}^{1} f\left(x\right)dx = \int_{-1}^{0} \left(x+1\right)dx + \int_{0}^{1} \left(x^{2} +1\right)dx
=[x22+x]10+[x3x+x]01= \left[\frac{x^{2}}{2}+x\right]^{0}_{1} + \left[\frac{x^{3}}{x} + x\right]^{1}_{0}
=−[12−1]+(13+1)=12+43=116= - \left[\frac{1}{2} -1\right]+ \left(\frac{1}{3} +1\right) = \frac{1}{2} + \frac{4}{3} = \frac{11}{6}