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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 $$$\int_{3}^{9}\left( \frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}} \right)dy$$$
First, calculate the corresponding indefinite integral: $$$\int{\left(\frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}}\right)d y}=\frac{\sqrt{y} \left(y + 3\right)}{3}$$$ (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.
$$$\left(\frac{\sqrt{y} \left(y + 3\right)}{3}\right)|_{\left(y=9\right)}=12$$$
$$$\left(\frac{\sqrt{y} \left(y + 3\right)}{3}\right)|_{\left(y=3\right)}=2 \sqrt{3}$$$
$$$\int_{3}^{9}\left( \frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}} \right)dy=\left(\frac{\sqrt{y} \left(y + 3\right)}{3}\right)|_{\left(y=9\right)}-\left(\frac{\sqrt{y} \left(y + 3\right)}{3}\right)|_{\left(y=3\right)}=12 - 2 \sqrt{3}$$$
Answer: $$$\int_{3}^{9}\left( \frac{\sqrt{y}}{2} + \frac{1}{2 \sqrt{y}} \right)dy=12 - 2 \sqrt{3}\approx 8.53589838486224$$$