The region is bounded by the graph of the function: f(x)=21+x2 x=-1, x=1 Tell the exact area.

∫−1121+x2dx ∫−1121+x2dx=2∫−1111+x2dx=4∫0111+x2dx=4(arctan⁡1−arctan⁡0)=4(π4−0)=π As f(x) is even

An even function is when f(x)=f(-x) and is visualized as a function that is symmetric over the x-axis. This means in terms of area, that ∫0111+x2dx=∫−1011+x2dx. Thus the problem is: the function is even, 2∫−1111+x2dx can more easily be found by finding 2(∫0111+x2dx+∫0111+x2dx) or 4∫0111+x2dx.

∫−1121+x2dx ∫−1121+x2dx=2∫−1111+x2dx=4∫0111+x2dx=4(arctan⁡1−arctan⁡0)=4(π4−0)=π As f(x) is even

An even function is when f(x)=f(-x) and is visualized as a function that is symmetric over the x-axis. This means in terms of area, that ∫0111+x2dx=∫−1011+x2dx. Thus the problem is: the function is even, 2∫−1111+x2dx can more easily be found by finding 2(∫0111+x2dx+∫0111+x2dx) or 4∫0111+x2dx.

Key to "The region is bounded by the graph of..."

Sam Longoria

Answered question

2021-12-10

The region is bounded by the graph of the function:

f (x) = \frac{2}{1 + x^{2}}

x=-1, x=1
Tell the exact area.

Answer & Explanation

Ella Williams

Beginner2021-12-11Added 28 answers

\int_{- 1}^{1} \frac{2}{1 + x^{2}} d x

\int_{- 1}^{1} \frac{2}{1 + x^{2}} d x = 2 \int_{- 1}^{1} \frac{1}{1 + x^{2}} d x = 4 \int_{0}^{1} \frac{1}{1 + x^{2}} d x = 4 (\arctan 1 - \arctan 0) = 4 (\frac{π}{4} - 0) = π

As f(x) is even

Mason Hall

Beginner2021-12-12Added 36 answers

An even function is when f(x)=f(-x) and is visualized as a function that is symmetric over the x-axis.
This means in terms of area, that

\int_{0}^{1} \frac{1}{1 + x^{2}} d x = \int_{- 1}^{0} \frac{1}{1 + x^{2}} d x

. Thus the problem is: the function is even,

2 \int_{- 1}^{1} \frac{1}{1 + x^{2}} d x

can more easily be found by finding

2 (\int_{0}^{1} \frac{1}{1 + x^{2}} d x + \int_{0}^{1} \frac{1}{1 + x^{2}} d x)

4 \int_{0}^{1} \frac{1}{1 + x^{2}} d x

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