Määrättyjen ja epäoleellisten integraalien laskin

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

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Solution

Your input: calculate $$$\int_{1}^{5}\left( w^{2} \ln{\left(w \right)} \right)dw$$$

First, calculate the corresponding indefinite integral: $$$\int{w^{2} \ln{\left(w \right)} d w}=\frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}$$$ (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.

$$$\left(\frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}\right)|_{\left(w=5\right)}=- \frac{125}{9} + \frac{125 \ln{\left(5 \right)}}{3}$$$

$$$\left(\frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}\right)|_{\left(w=1\right)}=- \frac{1}{9}$$$

$$$\int_{1}^{5}\left( w^{2} \ln{\left(w \right)} \right)dw=\left(\frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}\right)|_{\left(w=5\right)}-\left(\frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}\right)|_{\left(w=1\right)}=- \frac{124}{9} + \frac{125 \ln{\left(5 \right)}}{3}$$$

Answer: $$$\int_{1}^{5}\left( w^{2} \ln{\left(w \right)} \right)dw=- \frac{124}{9} + \frac{125 \ln{\left(5 \right)}}{3}\approx 53.2821352403097$$$

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Määrättyjen ja epäoleellisten integraalien laskin

Laske määrättyjä ja epäoleellisia integraaleja askel askeleelta

Solution