The shaded region is bounded by the graph of the function
$$
f(x)=\frac{1}{x^{2}+1} \text {, }
$$
the line $x=k$, and the two coordinate axes.

If the region has area $\pi / 3$, what is the exact value of $k$ ?

Question

The shaded region is bounded by the graph of the function
$$
f(x)=\frac{1}{x^{2}+1} \text {, }
$$
the line $x=k$, and the two coordinate axes.

If the region has area $\pi / 3$, what is the exact value of $k$ ?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We need to find the value of $k$ such that the area under the curve $f(x) = \frac{1}{x^2 + 1}$ from $x = 0$ to $x = k$ is $\frac{\pi}{3}$. This can be expressed as the definite integral: $$ \int_0^k \frac{1}{x^2 + 1} \, dx = \frac{\pi}{3} $$ ‖ ‖ step 2 ⋮ The integral of $\frac{1}{x^2 + 1}$ is $\arctan(x)$. Therefore, we can rewrite the integral as: $$ \left[ \arctan(x) \right]_0^k = \frac{\pi}{3} $$ ‖ ‖ step 3 ⋮ Evaluating the integral at the bounds, we get: $$ \arctan(k) - \arctan(0) = \frac{\pi}{3} $$ ‖ ‖ step 4 ⋮ Since $\arctan(0) = 0$, the equation simplifies to: $$ \arctan(k) = \frac{\pi}{3} $$ ‖ ‖ step 5 ⋮ Solving for $k$, we take the tangent of both sides: $$ k = \tan\left(\frac{\pi}{3}\right) $$ ‖ ‖ step 6 ⋮ We know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$, so: $$ k = \sqrt{3} $$ ‖ ∻Answer∻ ⚹ $k = \sqrt{3}$ ⚹ ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval gives the area under the curve of that function over that interval. In this problem, we used the integral of $\frac{1}{x^2 + 1}$ to find the value of $k$ that results in a specific area.⚹◊What is the formula for calculating the area under a curve bounded by the x-axis?⍭ Generate me a similar question◊

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