Määrättyjen ja epäoleellisten integraalien laskin

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

Enter a function:

Integrate with respect to:

Enter a lower limit:

If you need `-oo`, type -inf.

Enter an upper limit:

If you need `oo`, type inf.

Please write without any differentials such as `dx`, `dy` etc.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

Solution

Your input: calculate $$$\int_{0}^{\frac{\pi}{4}}\left( \cos^{3}{\left(2 x \right)} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{\cos^{3}{\left(2 x \right)} d x}=\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}$$$ (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.

$$$\left(\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}\right)|_{\left(x=\frac{\pi}{4}\right)}=\frac{1}{3}$$$

$$$\left(\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}\right)|_{\left(x=0\right)}=0$$$

$$$\int_{0}^{\frac{\pi}{4}}\left( \cos^{3}{\left(2 x \right)} \right)dx=\left(\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}\right)|_{\left(x=\frac{\pi}{4}\right)}-\left(\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}\right)|_{\left(x=0\right)}=\frac{1}{3}$$$

Answer: $$$\int_{0}^{\frac{\pi}{4}}\left( \cos^{3}{\left(2 x \right)} \right)dx=\frac{1}{3}\approx 0.333333333333333$$$

Please try a new game Rotatly

Määrättyjen ja epäoleellisten integraalien laskin

Laske määrättyjä ja epäoleellisia integraaleja askel askeleelta

Solution