The graph of function $g$ is shown below. Let $h(x)=\int_{-5}^{x} g(t) d t$.

Evaluate $h(0)$.
$$
h(0)=
$$

Question

The graph of function $g$ is shown below. Let $h(x)=\int_{-5}^{x} g(t) d t$.

Evaluate $h(0)$.
$$
h(0)=
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To evaluate $h(0)$, we need to compute the definite integral of $g(t)$ from $-5$ to $0$. This is given by $h(0) = \int_{-5}^{0} g(t) \, dt$. ‖ ‖ step 2 ⋮ From the graph, we can see that $g(t)$ forms a triangle from $(-5,0)$ to $(-3,1)$ and then another triangle from $(-3,1)$ to $(0,0)$. We need to find the area of these triangles. ‖ ‖ step 3 ⋮ The first triangle has a base of $2$ units (from $-5$ to $-3$) and a height of $1$ unit. The area of this triangle is $\frac{1}{2} imes 2 imes 1 = 1$. ‖ ‖ step 4 ⋮ The second triangle has a base of $3$ units (from $-3$ to $0$) and a height of $1$ unit. The area of this triangle is $\frac{1}{2} imes 3 imes 1 = 1.5$. ‖ ‖ step 5 ⋮ Since the graph of $g(t)$ is above the $t$-axis in this interval, the areas are positive. Therefore, the total area under $g(t)$ from $-5$ to $0$ is $1 + 1.5 = 2.5$. ‖ ‖ step 6 ⋮ Thus, $h(0) = 2.5$. ‖ ∻Answer∻ ⚹ $h(0) = 2.5$ ⚹ ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval can be interpreted as the area under the curve of the function within that interval.⚹◊What is the definition of the definite integral $h(x)=\int_{-5}^{x} g(t) dt$ in this context⍭ Generate me a similar question◊

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