問題文全文(内容文)：
次の定積分を求めよ。
(1) $\displaystyle \int_{-1}^{0}(x+2)\sqrt{3x+4}\,dx$
(2) $\displaystyle \int_{0}^{4}\dfrac{x^2}{\sqrt{x+1}}\,dx$
(3) $\displaystyle \int_{0}^{1}\dfrac{x^3}{\sqrt{1+x^2}}\,dx$
(4) $\displaystyle \int_{1}^{3}\dfrac{dx}{x\sqrt{x+1}}$
(5) $\displaystyle \int_{1}^{2}\dfrac{dx}{e^x-1}$
(6) $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{\sin^3x}{\cos^2x}\,dx$
次の定積分を求めよ。
(1) $\displaystyle \int_{-1}^{0}(x+2)\sqrt{3x+4}\,dx$
(2) $\displaystyle \int_{0}^{4}\dfrac{x^2}{\sqrt{x+1}}\,dx$
(3) $\displaystyle \int_{0}^{1}\dfrac{x^3}{\sqrt{1+x^2}}\,dx$
(4) $\displaystyle \int_{1}^{3}\dfrac{dx}{x\sqrt{x+1}}$
(5) $\displaystyle \int_{1}^{2}\dfrac{dx}{e^x-1}$
(6) $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{\sin^3x}{\cos^2x}\,dx$
単元：
#積分とその応用#定積分#数学(高校生)#数Ⅲ
教材：
#4S数学#4S数学ⅢのB問題解説#中高教材#積分法の応用
指導講師：
理数個別チャンネル
問題文全文(内容文)：
次の定積分を求めよ。
(1) $\displaystyle \int_{-1}^{0}(x+2)\sqrt{3x+4}\,dx$
(2) $\displaystyle \int_{0}^{4}\dfrac{x^2}{\sqrt{x+1}}\,dx$
(3) $\displaystyle \int_{0}^{1}\dfrac{x^3}{\sqrt{1+x^2}}\,dx$
(4) $\displaystyle \int_{1}^{3}\dfrac{dx}{x\sqrt{x+1}}$
(5) $\displaystyle \int_{1}^{2}\dfrac{dx}{e^x-1}$
(6) $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{\sin^3x}{\cos^2x}\,dx$
次の定積分を求めよ。
(1) $\displaystyle \int_{-1}^{0}(x+2)\sqrt{3x+4}\,dx$
(2) $\displaystyle \int_{0}^{4}\dfrac{x^2}{\sqrt{x+1}}\,dx$
(3) $\displaystyle \int_{0}^{1}\dfrac{x^3}{\sqrt{1+x^2}}\,dx$
(4) $\displaystyle \int_{1}^{3}\dfrac{dx}{x\sqrt{x+1}}$
(5) $\displaystyle \int_{1}^{2}\dfrac{dx}{e^x-1}$
(6) $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{\sin^3x}{\cos^2x}\,dx$
投稿日：2026.08.24
