Evaluate:∫(x→0 to 1) dx/1+x^2
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Answer:
π2.
Explanation:
We know that ∫11+x2dx=arctanx+C.
Then, by the Fundamental Principle of Definite Integral,
∴I=∫∞011+x2dx=limy→∞∫y011+x2dx
=limy→∞[arctanx]y0
=limy→∞[arctany−arctan0]
Since, the arctan fun. is continuous on R, we have,
I=arctan{limy→∞y}−0
=π2.
π2.
Explanation:
We know that ∫11+x2dx=arctanx+C.
Then, by the Fundamental Principle of Definite Integral,
∴I=∫∞011+x2dx=limy→∞∫y011+x2dx
=limy→∞[arctanx]y0
=limy→∞[arctany−arctan0]
Since, the arctan fun. is continuous on R, we have,
I=arctan{limy→∞y}−0
=π2.
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