The value of ∫−4−5 e(x+5)2dx+3∫1/32/3 e9(x−2/3)2dx , is

Let I1=∫−4−5 e(x+5)2dx and I2=∫1/32/3 e9(x−2/3)2dxUsing Property X, we haveI1=∫−4−5 e(x+5)2dx⇒ I1=(−5+4)∫01 e1(−5+4)x−4+5γ2dx⇒ I1=−∫01 e(−x+1)2dx=−∫01 e(x−1)2dx and, I2=∫1/32/3 e9(x−2/3)2dx⇒ I2 =23−13∫01 e9{(2/3−1/3)x+1/3−2/3}2dx⇒ I2 =13∫01 e9(x/3−1/3)2dx=13∫01 e(x−1)2dx∴ I =I1+3I2=−∫01 e(x−1)2dx+∫01 e(x−1)2dx=0

The value of ∫−4−5 e(x+5)2dx+3∫1/32/3 e9(x−2/3)2dx , is

Let I1=∫−4−5 e(x+5)2dx and I2=∫1/32/3 e9(x−2/3)2dxUsing Property X, we haveI1=∫−4−5 e(x+5)2dx⇒ I1=(−5+4)∫01 e1(−5+4)x−4+5γ2dx⇒ I1=−∫01 e(−x+1)2dx=−∫01 e(x−1)2dx and, I2=∫1/32/3 e9(x−2/3)2dx⇒ I2 =23−13∫01 e9{(2/3−1/3)x+1/3−2/3}2dx⇒ I2 =13∫01 e9(x/3−1/3)2dx=13∫01 e(x−1)2dx∴ I =I1+3I2=−∫01 e(x−1)2dx+∫01 e(x−1)2dx=0

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The value of $\int_{- 4}^{- 5} e^{(x + 5)^{2}} d x + 3 \int_{1 / 3}^{2 / 3} e^{9 (x - 2 / 3)^{2}} d x$ , is

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Detailed Solution

Let $I_{1} = \int_{- 4}^{- 5} e^{(x + 5)^{2}} d x and I_{2} = \int_{1 / 3}^{2 / 3} e^{9 (x - 2 / 3)^{2}} d x$

Using Property X, we have

$\begin{array}{l} I_{1} = \int_{- 4}^{- 5} e^{(x + 5)^{2}} d x \\ \Rightarrow I_{1} = (- 5 + 4) \int_{0}^{1} e^{1 (- 5 + 4) x - 4 + 5 γ^{2}} d x \\ \Rightarrow I_{1} = - \int_{0}^{1} e^{(- x + 1)^{2}} d x = - \int_{0}^{1} e^{(x - 1)^{2}} d x \\ and, I_{2} = \int_{1 / 3}^{2 / 3} e^{9 (x - 2 / 3)^{2}} d x \end{array}$

$\begin{array}{ll} \Rightarrow I_{2} = (\frac{2}{3} - \frac{1}{3}) \int_{0}^{1} e^{9 {(2 / 3 - 1 / 3) x + 1 / 3 - 2 / 3}^{2}} d x \\ \Rightarrow I_{2} = \frac{1}{3} \int_{0}^{1} e^{9 (x / 3 - 1 / 3)^{2}} d x = \frac{1}{3} \int_{0}^{1} e^{(x - 1)^{2}} d x \\ ∴ I = I_{1} + 3 I_{2} = - \int_{0}^{1} e^{(x - 1)^{2}} d x + \int_{0}^{1} e^{(x - 1)^{2}} d x = 0 \end{array}$