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

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

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Find derivatives of sine and cosine functions

Full solution

Q.

g

be the function given by

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

. Find

g(1)

.

\newline

f_{0} 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 limits of integration: : Identify the limits of integration for $g(1)$ . $g(1) = \int_{-1}^{1}f(t)\,dt$
Split integral at point: : Split the integral at the point where the definition of $f(t)$ changes, which is at $t = 0$ . $g(1) = \int_{-1}^{0}(3t^2 + 2t)dt + \int_{0}^{1}(e^{2t} + 2)dt$
Calculate first integral: : Calculate the first integral from $-1$ to $0$ . $\int_{-1}^{0}(3t^2 + 2t)dt = [t^3 + t^2]_{-1}^{0}$ $= (0^3 + 0^2) - ((-1)^3 + (-1)^2)$ $= 0 - (-1 + 1)$ $= 0$

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