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$Let g be the function given by g(x)=int_(-1)^(x)f(t)dt. Find g(1 fos={[3x^(2)+2x," for ",x <= 0],[e^(2x)+2," for ",x > 0]:}$

Let $g$ be the function given by $g(x)=\int_{-1}^{x} f(t) d t$ . Find $g(1$ $\newline$ $f o s=\left\{\begin{array}{lll} 3 x^{2}+2 x & \text { for } & x \leq 0 \\ e^{2 x}+2 & \text { for } & x>0 \end{array}\right.$

Full solution

Q. Let

g

be the function given by

g(x)=\int_{-1}^{x} f(t) d t

. Find

g(1

\newline

f o s=\left\{\begin{array}{lll} 3 x^{2}+2 x & \text { for } & x \leq 0 \\ e^{2 x}+2 & \text { for } & x>0 \end{array}\right.

Identify function $f(t)$ : : Identify the function $f(t)$ for the interval from $-1$ to $1$ . $\newline$ $f(t)$ is defined as $3t^2 + 2t$ for $t \leq 0$ and $e^{(2t)} + 2$ for $t > 0$ .
Split integral at $t=0$ : : Split the integral at the point where the function definition changes, which is at $t = 0$ . $g(1) = \int_{-1}^{0}(3t^2 + 2t)\,dt + \int_{0}^{1}(e^{2t} + 2)\,dt$ .
Calculate integral from $-1$ to $0$ : : Calculate the first integral from $-1$ to $0$ . $\int_{-1}^{0}(3t^2 + 2t)dt = [t^3 + t^2]_{-1}^{0} = (0^3 + 0^2) - (-1^3 + -1^2) = 0 - (-1 + 1) = 0.$
Calculate integral from $0$ to $1$ : : Calculate the second integral from $0$ to $1$ . $\int_{0}^{1}(e^{2t} + 2)dt = \left[\frac{e^{2t}}{2} + 2t\right]$ from $0$ to $1$ = $\left(\frac{e^{2\cdot 1}}{2} + 2\cdot 1\right) - \left(\frac{e^{2\cdot 0}}{2} + 2\cdot 0\right) = \left(\frac{e^2}{2} + 2\right) - \left(\frac{1}{2} + 0\right)$ .
Combine results of integrals: : Combine the results of the two integrals. $g(1) = 0 + (\frac{e^2}{2} + 2) - (\frac{1}{2} + 0) = \frac{e^2}{2} + 2 - \frac{1}{2}$ .
Simplify expression: : Simplify the expression. $g(1) = \frac{e^2}{2} + \frac{3}{2}$ .