定積分・広義積分計算機
定積分と広義積分を手順を追って計算する
この計算機は、定積分（すなわち上下限のある積分）を、広義積分も含めて、解法手順を示しながら計算を試みます。
Solution
Your input: calculate $$$\int_{0}^{1}\left( t^{2} x - t^{2} y \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{\left(t^{2} x - t^{2} y\right)d x}=\frac{t^{2} x \left(x - 2 y\right)}{2}$$$ (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.
$$$\left(\frac{t^{2} x \left(x - 2 y\right)}{2}\right)|_{\left(x=1\right)}=\frac{t^{2} \left(1 - 2 y\right)}{2}$$$
$$$\left(\frac{t^{2} x \left(x - 2 y\right)}{2}\right)|_{\left(x=0\right)}=0$$$
$$$\int_{0}^{1}\left( t^{2} x - t^{2} y \right)dx=\left(\frac{t^{2} x \left(x - 2 y\right)}{2}\right)|_{\left(x=1\right)}-\left(\frac{t^{2} x \left(x - 2 y\right)}{2}\right)|_{\left(x=0\right)}=\frac{t^{2} \left(1 - 2 y\right)}{2}$$$
Answer: $$$\int_{0}^{1}\left( t^{2} x - t^{2} y \right)dx=\frac{t^{2} \left(1 - 2 y\right)}{2}$$$