다음을 풀어보세요: ∫ (from - 1 to 1) of e^xdu=

계산

2e^{x}

해답 단계

\int _ { - 1 } ^ { 1 } e ^ { x } d u =

먼저 부정적분을 구합니다.

\int e^{x}\mathrm{d}u

일반 적분 규칙 \int a\mathrm{d}u=au 표를 사용 하 여 e^{x}의 적분을 구합니다.

e^{x}u

정적분은 적분의 상한에서 구해진 수식의 미분 계수에서 적분의 하한에서 계산된 미분 계수를 뺀 값입니다.

e^{x}\times 1-e^{x}\left(-1\right)

단순화합니다.

2e^{x}

x 관련 미분

2e^{x}

퀴즈

Integration

\int _ { - 1 } ^ { 1 } e ^ { x } d u =

비슷한 문제의 웹 검색 결과

taylor expansion of an integral \int_0^1{e^{x^2}}

https://math.stackexchange.com/q/379081

You should use a different form of the remainder term, namely the tail of the series: R_n(x)=\sum_{k=n+1}^\infty \frac{x^{2k}}{k!} and then you estimate that, for example by comparing with a ...

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https://math.stackexchange.com/questions/2973300/int-x1-et3-dt

Wrong way to go. \int_{x}^{1}e^{t^3}\,dt , just like \int_{0}^{a}e^{-x^2}\,dx , is not an elementary function. On the other hand by Cavalieri's principle \iiint_{0\leq y\leq x\leq t\leq 1}e^{t^3}\,dt\,dx\,dy = \int_{0}^{1} \frac{t^2}{2}e^{t^3}\,dt=\left[\frac{1}{6}e^{t^3}\right]_{0}^{1}=\color{red}{\frac{e-1}{6}}.

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https://math.stackexchange.com/questions/2778076/word-integral-problem-check-did-i-do-this-correctly

The initial set up is correct, by the change of variable we have u=e^x\implies du=e^x dx and then \int_{-1}^1 \pi(e^x)^2\, dx=\int_{-1}^1 \pi e^x e^x\, dx=\int_{1/e}^e \pi u\, du=\pi[u^2/2]_{1/e}^e=\frac \pi 2\left(e^2 -\frac1 {e^2}\right)

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https://math.stackexchange.com/questions/1229764/bounds-on-maclaurin-series-of-e-x2

The reason is that your sequence 1 - x^2 + \frac{x^4}{2!}- \frac{x^6}{3!} + .... = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!} is alternating. In this case, if you use the first n-th term to ...

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https://math.stackexchange.com/questions/850287/calculate-integrals-using-euler-maclaurins-formula

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https://math.stackexchange.com/q/2736793

You have a typo at \int_0^1 e^x dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{n} e^{\frac{k}{n}} = \color{red}{ \lim_{n\rightarrow \infty} \frac{1-e}{n-ne^n}} It should be \int_0^1 e^x dx = \lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{n} e^{\frac{k}{n}} = \color{blue}{ \lim_{n\rightarrow \infty} \frac{(1 - e) e^{\frac{1}{n}}}{n\left(1 - e^{\frac{1}{n}}\right)}} ...

클립보드에 복사됨

\int e^{x}\mathrm{d}u

먼저 부정적분을 구합니다.

e^{x}u

일반 적분 규칙 \int a\mathrm{d}u=au 표를 사용 하 여 e^{x}의 적분을 구합니다.

e^{x}\times 1-e^{x}\left(-1\right)

정적분은 적분의 상한에서 구해진 수식의 미분 계수에서 적분의 하한에서 계산된 미분 계수를 뺀 값입니다.

2e^{x}

단순화합니다.

예제

이차방정식

{ x } ^ { 2 } - 4 x - 5 = 0

삼각법

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

연립방정식

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

미분

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

적분

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

극한

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

주제

주제

비슷한 문제의 웹 검색 결과

공유

예제