以下の定積分を計算します。 $\int_{1}^{4} \left( \sqrt[4]{x^3} + \frac{1}{\sqrt{2x}} + (3x)^2 \right) dx$

1. 問題の内容

以下の定積分を計算します。

$\int_{1}^{4} \left( \sqrt[4]{x^3} + \frac{1}{\sqrt{2x}} + (3x)^2 \right) dx$

2. 解き方の手順

まず、積分を分割して、各項を個別に積分します。

$\int_{1}^{4} \sqrt[4]{x^3} dx + \int_{1}^{4} \frac{1}{\sqrt{2x}} dx + \int_{1}^{4} (3x)^2 dx$

それぞれの積分を計算します。

$\int_{1}^{4} \sqrt[4]{x^3} dx = \int_{1}^{4} x^{\frac{3}{4}} dx = \left[ \frac{4}{7} x^{\frac{7}{4}} \right]_{1}^{4} = \frac{4}{7} (4^{\frac{7}{4}} - 1^{\frac{7}{4}}) = \frac{4}{7} (2^2)^{\frac{7}{4}} - \frac{4}{7} = \frac{4}{7} (2^{\frac{14}{4}} - 1) = \frac{4}{7} (2^{\frac{7}{2}} - 1) = \frac{4}{7} (2^3 \sqrt{2} - 1) = \frac{4}{7} (8\sqrt{2} - 1) = \frac{32\sqrt{2} - 4}{7}$

$\int_{1}^{4} \frac{1}{\sqrt{2x}} dx = \frac{1}{\sqrt{2}} \int_{1}^{4} x^{-\frac{1}{2}} dx = \frac{1}{\sqrt{2}} \left[ 2x^{\frac{1}{2}} \right]_{1}^{4} = \frac{2}{\sqrt{2}} (\sqrt{4} - \sqrt{1}) = \frac{2}{\sqrt{2}} (2-1) = \frac{2}{\sqrt{2}} = \sqrt{2}$

$\int_{1}^{4} (3x)^2 dx = \int_{1}^{4} 9x^2 dx = \left[ 3x^3 \right]_{1}^{4} = 3(4^3 - 1^3) = 3(64 - 1) = 3(63) = 189$

したがって、元の積分は

$\frac{32\sqrt{2} - 4}{7} + \sqrt{2} + 189 = \frac{32\sqrt{2} - 4 + 7\sqrt{2}}{7} + 189 = \frac{39\sqrt{2} - 4}{7} + 189 = \frac{39\sqrt{2} - 4 + 189(7)}{7} = \frac{39\sqrt{2} - 4 + 1323}{7} = \frac{39\sqrt{2} + 1319}{7}$

3. 最終的な答え

$\frac{39\sqrt{2} + 1319}{7}$