Määrättyjen ja epäoleellisten integraalien laskin

Laskin yrittää laskea määrätyn (eli rajoilla varustetun) integraalin, mukaan lukien epäoleelliset tapaukset, ja näyttää välivaiheet.

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Integrate with respect to:

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Solution

Your input: calculate $$$\int_{0}^{1}\left( \sqrt{1 - y^{2}} \right)dy$$$

First, calculate the corresponding indefinite integral: $$$\int{\sqrt{1 - y^{2}} d y}=\frac{y \sqrt{1 - y^{2}}}{2} + \frac{\operatorname{asin}{\left(y \right)}}{2}$$$ (for steps, see indefinite integral calculator)

According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.

$$$\left(\frac{y \sqrt{1 - y^{2}}}{2} + \frac{\operatorname{asin}{\left(y \right)}}{2}\right)|_{\left(y=1\right)}=\frac{\pi}{4}$$$

$$$\left(\frac{y \sqrt{1 - y^{2}}}{2} + \frac{\operatorname{asin}{\left(y \right)}}{2}\right)|_{\left(y=0\right)}=0$$$

$$$\int_{0}^{1}\left( \sqrt{1 - y^{2}} \right)dy=\left(\frac{y \sqrt{1 - y^{2}}}{2} + \frac{\operatorname{asin}{\left(y \right)}}{2}\right)|_{\left(y=1\right)}-\left(\frac{y \sqrt{1 - y^{2}}}{2} + \frac{\operatorname{asin}{\left(y \right)}}{2}\right)|_{\left(y=0\right)}=\frac{\pi}{4}$$$

Answer: $$$\int_{0}^{1}\left( \sqrt{1 - y^{2}} \right)dy=\frac{\pi}{4}\approx 0.785398163397448$$$

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Määrättyjen ja epäoleellisten integraalien laskin

Laske määrättyjä ja epäoleellisia integraaleja askel askeleelta

Solution