Let f, f', f'' be continuous in [0, ln 2] and | vedclass

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(B) We use integration by parts for the integral $I = \int_{0}^{\ln 2} e^{-2x} f''(x) dx$.Let $u = e^{-2x}$ and $dv = f''(x) dx$. Then $du = -2e^{-2x}

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$13$

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(B) We use integration by parts for the integral $I = \int_{0}^{\ln 2} e^{-2x} f''(x) dx$.Let $u = e^{-2x}$ and $dv = f''(x) dx$. Then $du = -2e^{-2x} dx$ and $v = f'(x)$.Using $\int u dv = uv - \int v du$,we get:$I = [e^{-2x} f'(x)]_{0}^{\ln 2} - \int_{0}^{\ln 2} f'(x) (-2e^{-2x}) dx$$I = [e^{-2\ln 2} f'(\ln 2) - e^{0} f'(0)] + 2 \int_{0}^{\ln 2} e^{-2x} f'(x) dx$Since $e^{-2\ln 2} = e^{\ln(2^{-2})} = \frac{1}{4}$,we have:$I = [\frac{1}{4}(4) - 1(3)] + 2 \int_{0}^{\ln 2} e^{-2x} f'(x) dx = (1 - 3) + 2 \int_{0}^{\ln 2} e^{-2x} f'(x) dx = -2 + 2 \int_{0}^{\ln 2} e^{-2x} f'(x) dx$.Now,integrate $\int_{0}^{\ln 2} e^{-2x} f'(x) dx$ by parts again:Let $u = e^{-2x}$ and $dv = f'(x) dx$. Then $du = -2e^{-2x} dx$ and $v = f(x)$.$\int_{0}^{\ln 2} e^{-2x} f'(x) dx = [e^{-2x} f(x)]_{0}^{\ln 2} - \int_{0}^{\ln 2} f(x) (-2e^{-2x}) dx$$= [e^{-2\ln 2} f(\ln 2) - e^{0} f(0)] + 2 \int_{0}^{\ln 2} e^{-2x} f(x) dx$$= [\frac{1}{4}(6) - 0] + 2(3) = 1.5 + 6 = 7.5$.Substituting this back into the expression for $I$:$I = -2 + 2(7.5) = -2 + 15 = 13$.

Let $f, f', f''$ be continuous in $[0, \ln 2]$ and $f(0) = 0, f'(0) = 3, f(\ln 2) = 6, f'(\ln 2) = 4$ and $\int_{0}^{\ln 2} e^{-2x} f(x) dx = 3$,then $\int_{0}^{\ln 2} e^{-2x} f''(x) dx$ is

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