Graph $y=2 \cos \left(\frac{1}{2} x\right)-1$ in the interactive widget.
Note that one moveable point always defines an extremum point in the graph and the other point always defines a neighbouring intersection with the midline.

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∻Solution by Steps∻ ‖ step 1 ⋮ To evaluate the definite integral, we use the given function under the square root. ‖ ‖ step 2 ⋮ The function to integrate is $\sqrt{16 e^{-2 t} + \sin^2(1 + \sqrt{t})}$. ‖ ‖ step 3 ⋮ We integrate the function from $t=1$ to $t=3$. ‖ ‖ step 4 ⋮ The result of the integration is approximately $1.8606$. ‖ ∻1 Answer∻ ⚹ The integral of the given function from $t=1$ to $t=3$ is approximately $1.8606$. ⚹ ∻Key Concept∻ ⚹Definite Integration of Functions⚹ ∻Explanation∻ ⚹The definite integral of a function gives the area under the curve of the function between two points on the x-axis. In this case, the function is a combination of an exponential decay and a squared sine function, integrated from $t=1$ to $t=3$.⚹◊What is the period of the cosine function $y = 2 \cos\left(\frac{1}{2}x\right) - 1$⍭ Generate me a similar question◊

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