Primitive

Exerciții și probleme... primitive.

Matematică >> primitive >> 1


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Mulțimea primitivelor
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( f(x) = x^n \),
unde \( n \in \mathbb{N} \)
\( \displaystyle \int x^n dx = \) \( \displaystyle \frac{x^{n+1}}{n+1} +C \)
\( f : I \rightarrow \mathbb{R} \), \( f(x) = x^{\alpha} \),
unde \( I \subset (0, \infty) \), \( \alpha \in \mathbb{R} \setminus \{ -1 \} \)
\( \displaystyle \int x^{\alpha} dx = \) \( \displaystyle \frac{x^{\alpha+1}}{\alpha+1} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{x} \),
unde \( I \subset (0, \infty) \) sau \( I \subset (-\infty, 0) \)
\( \displaystyle \int \frac{1}{x} dx = \) \( \displaystyle \ln{ \mid x \mid } +C \)
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( f(x) = e^x \)
\( \displaystyle \int e^x dx = \) \( \displaystyle e^x +C \)
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( f(x) = a^x \),
unde \( a \gt 0 \), \( a \ne 1 \)
\( \displaystyle \int a^x dx = \) \( \displaystyle \frac{a^x}{ \ln {a}} +C \)
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{x^2 + a^2} \),
unde \( a \ne 0 \)
\( \displaystyle \int \frac{1}{x^2 + a^2} dx = \) \( \displaystyle \frac{1}{a}\text{arctg}{ \frac{x}{a} } +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{x^2 - a^2} \),
unde \( I \subset \mathbb{R} \setminus \{ -a, a \} \), \( a \ne 0 \)
\( \displaystyle \int \frac{1}{x^2 - a^2} dx = \) \( \displaystyle \frac{1}{2a}\ln{ \left| \frac{x-a}{x+a} \right| } +C \)
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( f(x) = \sin{x} \)
\( \displaystyle \int \sin{x} dx = \) \( \displaystyle -\cos{x} +C \)
\( f : \mathbb{R} \rightarrow \mathbb{R} \), \( f(x) = \cos{x} \)
\( \displaystyle \int \cos{x} dx = \) \( \displaystyle \sin{x} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{\cos^2{x}} \),
unde \( \displaystyle I \subset \mathbb{R} \setminus \) \( \displaystyle \left\{ \frac{ (2k+1) \pi}{2} | k \in \mathbb{Z} \right\} \)
\( \displaystyle \int \frac{1}{\cos^2{x}} dx = \) \( \displaystyle \text{tg}{x} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{\sin^2{x}} \),
unde \( \displaystyle I \subset \mathbb{R} \setminus \) \( \displaystyle \left\{ k \pi | k \in \mathbb{Z} \right\} \)
\( \displaystyle \int \frac{1}{\sin^2{x}} dx = \) \( \displaystyle -\text{ctg}{x} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \text{tg}{x} \),
unde \( \displaystyle I \subset \mathbb{R} \setminus \) \( \displaystyle \left\{ \frac{ (2k+1) \pi}{2} | k \in \mathbb{Z} \right\} \)
\( \displaystyle \int \text{tg}{x} dx = \) \( \displaystyle -\ln{\left| \cos{x} \right|} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \text{ctg}{x} \),
unde \( \displaystyle I \subset \mathbb{R} \setminus \) \( \displaystyle \left\{ k \pi | k \in \mathbb{Z} \right\} \)
\( \displaystyle \int \text{ctg}{x} dx = \) \( \displaystyle \ln{\left| \sin{x} \right|} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{ \sqrt{x^2 + a^2} } \),
unde \( a \ne 0 \)
\( \displaystyle \int \frac{1}{ \sqrt{x^2 + a^2} } dx = \) \( \displaystyle \ln{\left( x + \sqrt{x^2+a^2} \right)} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{ \sqrt{x^2 - a^2} } \),
unde \( I \subset (-\infty, -a) \) sau \( I \subset (a, \infty) \), \( a \ne 0 \)
\( \displaystyle \int \frac{1}{ \sqrt{ x^2 - a^2 } } dx = \) \( \displaystyle \ln{\left| x + \sqrt{x^2-a^2} \right|} +C \)
\( f : I \rightarrow \mathbb{R} \), \( \displaystyle f(x) = \frac{1}{ \sqrt{a^2 - x^2} } \),
unde \( a \gt 0 \), \( I \subset (-a, a) \)
\( \displaystyle \int \frac{1}{ \sqrt{ a^2 - x^2 } } dx = \) \( \displaystyle \arcsin{ \frac{x}{a} } +C \)



exerciții

Calculați \( \displaystyle \int \cos{x} dx \), \( x \in \mathbb{R} \).

  \( \displaystyle \int \cos{x} dx = \) \( \displaystyle -\sin{x} +C \), \( x \in \mathbb{R} \)

  \( \displaystyle \int \cos{x} dx = \) \( \displaystyle \sin{x} +C \), \( x \in \mathbb{R} \)

  \( \displaystyle \int \cos{x} dx = \) \( \displaystyle e^{x} +C \), \( x \in \mathbb{R} \)

  \( \displaystyle \int \cos{x} dx = \) \( \displaystyle -\cos{x} +C \), \( x \in \mathbb{R} \)

  \( \displaystyle \int \cos{x} dx = \) \( \displaystyle \cos{x} +C \), \( x \in \mathbb{R} \)



 


exercițiu nou

\( \displaystyle \int \cos{x} dx = \) \( \displaystyle \sin{x} +C \), \( x \in \mathbb{R} \).

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