Question

If \(\mathrm{f}(\mathrm{x})=\left|2^{\mathrm{x}}-1\right|+|\mathrm{x}-1|\) then evaluate \(\int_{-2}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)

Short answer

The integral of f(x) = |2^x - 1| + |x - 1| on the interval [-2, 2] is 2 + 12/ln(2).

Step-by-step solution

Identify critical points

To find the critical points, we will solve for the values of \(x\) where either \(2^{\mathrm{x}}-1=0\) or \(\mathrm{x}-1=0\). For the first term, \(2^{\mathrm{x}}-1=0\), we get \(2^{\mathrm{x}}=1\). This happens only when \(x=0\). For the second term, \(\mathrm{x}-1=0\), we get \(x=1\). So, there are two critical points within the interval \([-2, 2]\): \(x=0\) and \(x=1\). These points divide the interval \([-2, 2]\) into three subintervals: \([-2, 0]\), \([0, 1]\), and \([1, 2]\).

Rewrite f(x) without absolute values on each subinterval

For \(x \in [-2, 0]\), both \(2^{\mathrm{x}}-1\) and \(\mathrm{x}-1\) are negative, so we can rewrite \(\mathrm{f}(\mathrm{x})\) as \(-\left(2^{\mathrm{x}}-1\right)-(\mathrm{x}-1)\). For \(x \in [0, 1]\), \(|\mathrm{x}-1|\) still needs to be positive, so we'll rewrite the function on this interval as \(-\left(2^{\mathrm{x}}-1\right)+(\mathrm{x}-1)\). For \(x \in [1, 2]\), both \(2^{\mathrm{x}}-1\) and \(\mathrm{x}-1\) are positive, so the function remains as \(\left(2^{\mathrm{x}}-1\right)+(\mathrm{x}-1)\).

Integrate f(x) on each subinterval

Now we will integrate \(\mathrm{f}(\mathrm{x})\) on each subinterval as follows: For \(x \in [-2, 0]\): \(\int_{-2}^{0}[\mathrm{-}(2^{\mathrm{x}}-1)-(\mathrm{x}-1)]\mathrm{d}\mathrm{x} = -\int_{-2}^{0}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{-2}^{0}\mathrm{x}\mathrm{d}\mathrm{x} -[-2,0]\). For \(x \in [0, 1]\): \(\int_{0}^{1}[\mathrm{-}(2^{\mathrm{x}}-1)+(\mathrm{x}-1)]\mathrm{d}\mathrm{x} = -\int_{0}^{1}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{0}^{1}\mathrm{x}\mathrm{d}\mathrm{x} -[0,1]\). For \(x \in [1, 2]\): \(\int_{1}^{2}[\mathrm{(2^{\mathrm{x}}-1)+(\mathrm{x}-1)}]\mathrm{d}\mathrm{x} = \int_{1}^{2}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{1}^{2}\mathrm{x}\mathrm{d}\mathrm{x} -[1,2]\).

Calculate the definite integrals on each subinterval and sum them up

Now, we will compute the resulting integrals in each segment: For \(x \in [-2, 0]\): \(-\int_{-2}^{0}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{-2}^{0}\mathrm{x}\mathrm{d}\mathrm{x} -[-2,0] = -[\frac{2^x}{\ln(2)}\Big|_{-2}^{0}] + [\frac{x^2}{2}\Big|_{-2}^{0}] - (0 - (-2))\) $= -\frac{1 - 2^{-2}}{\ln(2)} + \frac{4}{2} - 2 = \frac{11}{2\ln(2)} For \(x \in [0, 1]\): \(-\int_{0}^{1}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{0}^{1}\mathrm{x}\mathrm{d}\mathrm{x} -[0,1] = -[\frac{2^x}{\ln(2)}\Big|_{0}^{1}] + [\frac{x^2}{2}\Big|_{0}^{1}] - (1 - 0)\) \(= -\frac{2 - 1}{\ln(2)} + \frac{1}{2} - 1 = \frac{1}{2} - \frac{1}{\ln(2)}\) For \(x \in [1, 2]\): \(\int_{1}^{2}(2^{\mathrm{x}})\mathrm{d}\mathrm{x} +\int_{1}^{2}\mathrm{x}\mathrm{d}\mathrm{x} -[1,2] = [\frac{2^x}{\ln(2)}\Big|_{1}^{2}] + [\frac{x^2}{2}\Big|_{1}^{2}] - (2 - 1)\) \(= \frac{4 - 2}{\ln(2)} + \frac{3}{2} - 1 = \frac{3}{2} + \frac{2}{\ln(2)}\) Now, sum up the results of the three subintervals: \(\frac{11}{2\ln(2)}+\left(\frac{1}{2} - \frac{1}{\ln(2)}\right)+\left(\frac{3}{2} + \frac{2}{\ln(2)}\right) = 2+\frac{12}{\ln(2)}\) Thus, the definite integral of \(\mathrm{f}(\mathrm{x})\) over the interval \([-2, 2]\) is: \(\int_{-2}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}=2+\frac{12}{\ln(2)}\)