復習問題
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解答
$$\int f(t)dt=「微分がf(t)となる関数」+C\qquad(Cは積分定数)$$
- $\displaystyle\int t^2dt=\qquad$
- $\displaystyle\int \dfrac{1}{t^2}dt=\qquad$
- $\displaystyle\int \sqrt{t}dt=\qquad$
- $\displaystyle\int \dfrac{1}{\sqrt{t}}dt=\qquad$
- $\displaystyle\int \dfrac{1}{t}dt=\qquad$
- $\displaystyle\int (t^{\frac{3}{2}}-3t^{\frac{1}{2}})dt=\qquad$
- $\displaystyle\int \sin tdt=\qquad$
- $\displaystyle\int \dfrac{1+\cos t}{2}dt=\qquad$
- $\displaystyle\int \dfrac{1}{1+t^2}dt=\qquad$
- $\displaystyle\int \dfrac{1}{\sqrt{1-t^2}}dt=\qquad$
- $\displaystyle\int e^tdt=\qquad$