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1. (a) Use integration by parts to compute ∫12t2ln⁡tdt. \int_{1}^{2} t^{2} \ln t d t .
Jun 1, 2024
Solution by Steps
step 1
To solve the integral ∫12t2ln⁡t dt\int_{1}^{2} t^{2} \ln t \, dt using integration by parts, we first identify parts of the integrand to set uu and dvdv. Let u=ln⁡tu = \ln t and dv=t2 dtdv = t^2 \, dt
step 2
Compute dudu and vv. We have du=1t dtdu = \frac{1}{t} \, dt and v=∫t2 dt=t33v = \int t^2 \, dt = \frac{t^3}{3}
step 3
Apply the integration by parts formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Substituting the values, we get: ∫12t2ln⁡t dt=ln⁡t⋅t33∣12−∫12t33⋅1t dt. \int_{1}^{2} t^{2} \ln t \, dt = \left. \ln t \cdot \frac{t^3}{3} \right|_{1}^{2} - \int_{1}^{2} \frac{t^3}{3} \cdot \frac{1}{t} \, dt.
step 4
Simplify the expression: t3ln⁡t3∣12−∫12t23 dt. \left. \frac{t^3 \ln t}{3} \right|_{1}^{2} - \int_{1}^{2} \frac{t^2}{3} \, dt. Evaluate the first term: (23ln⁡23−13ln⁡13)=8ln⁡23−0=8ln⁡23. \left( \frac{2^3 \ln 2}{3} - \frac{1^3 \ln 1}{3} \right) = \frac{8 \ln 2}{3} - 0 = \frac{8 \ln 2}{3}.
step 5
Evaluate the second integral: ∫12t23 dt=13∫12t2 dt=13t33∣12=13(233−133)=13(83−13)=13⋅73=79. \int_{1}^{2} \frac{t^2}{3} \, dt = \frac{1}{3} \int_{1}^{2} t^2 \, dt = \frac{1}{3} \left. \frac{t^3}{3} \right|_{1}^{2} = \frac{1}{3} \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = \frac{1}{3} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{1}{3} \cdot \frac{7}{3} = \frac{7}{9}.
step 6
Combine the results: 8ln⁡23−79=24ln⁡29−79=24ln⁡2−79. \frac{8 \ln 2}{3} - \frac{7}{9} = \frac{24 \ln 2}{9} - \frac{7}{9} = \frac{24 \ln 2 - 7}{9}.
Answer
24ln⁡2−79\frac{24 \ln 2 - 7}{9}
Key Concept
Integration by Parts
Explanation
Integration by parts is a technique used to integrate products of functions by transforming the original integral into a simpler one.
怎么得到的，用了什么积分公式
Solution by Steps
step 1
To solve the integral ∫xex dx\int x e^{x} \, dx using integration by parts, we first identify parts of the integrand to set uu and dvdv. Let u=xu = x and dv=ex dxdv = e^x \, dx
step 2
Compute dudu and vv. We have du=dxdu = dx and v=∫ex dx=exv = \int e^x \, dx = e^x
step 3
Apply the integration by parts formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Substituting the values, we get ∫xex dx=xex−∫ex dx\int x e^x \, dx = x e^x - \int e^x \, dx
step 4
Simplify the integral. We have ∫xex dx=xex−ex+C\int x e^x \, dx = x e^x - e^x + C, where CC is the constant of integration
Answer
xex−ex+Cx e^x - e^x + C
Key Concept
Integration by Parts
Explanation
Integration by parts is a technique used to integrate products of functions by reducing them to simpler integrals.
为什么v=∫e x dx=e x
Solution by Steps
step 1
To solve the integral ∫ex dx\int e^x \, dx, we recognize that the integral of exe^x with respect to xx is exe^x
step 2
Therefore, ∫ex dx=ex+C\int e^x \, dx = e^x + C, where CC is the constant of integration
Answer
ex+Ce^x + C
Key Concept
Integration of exponential functions
Explanation
The integral of exe^x with respect to xx is exe^x plus a constant of integration, CC. This is a fundamental result in calculus.
fundamental result in calculus有哪些
Solution by Steps
step 1
The first fundamental theorem of calculus states that if ff is continuous on the closed interval [a,b][a, b] and FF is the antiderivative (indefinite integral) of ff on [a,b][a, b], then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a)
step 2
To apply this theorem, identify the function f(x)f(x) and its antiderivative F(x)F(x)
step 3
Evaluate F(x)F(x) at the upper limit bb and the lower limit aa
step 4
Subtract F(a)F(a) from F(b)F(b) to find the value of the definite integral ∫abf(x) dx\int_a^b f(x) \, dx
Answer
The first fundamental theorem of calculus states that ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a), where FF is the antiderivative of ff.
Key Concept
First Fundamental Theorem of Calculus
Explanation
This theorem connects the concept of differentiation and integration, showing that the definite integral of a function over an interval can be computed using its antiderivative.
把刚刚的解释翻译成中文
我会把刚刚的解释翻译成中文。在微积分中，通过分部积分法可以解决一些复杂的积分问题。同时，微积分基本定理是微积分中的重要概念，它将积分与导数联系起来，帮助我们计算积分。希望这些解释能帮助你更好地理解微积分的基本概念。
求积分的常见公式都有什么
Solution by Steps
step 1
常见的积分公式包括以下几种：
step 2
∫1 dx=x+C\int 1 \, dx = x + C
step 3
∫xn dx=xn+1n+1+C(n≠−1)\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)
step 4
∫ex dx=ex+C\int e^x \, dx = e^x + C
step 5
∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C
step 6
∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C
step 7
∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C
step 8
∫sec⁡2x dx=tan⁡x+C\int \sec^2 x \, dx = \tan x + C
step 9
∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x \, dx = -\cot x + C
step 10
∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x \, dx = \sec x + C
step 11
∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x \, dx = -\csc x + C
Answer
常见的积分公式如上所示
Key Concept
常见积分公式
Explanation
这些公式是积分计算中最基本和常用的公式，掌握这些公式可以帮助解决大多数基本积分问题。
Solution by Steps
step 2
The function cosh⁡(x)\cosh(x) is defined as cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^x + e^{-x}}{2}
step 3
Suppose cosh⁡(x1)=cosh⁡(x2)\cosh(x_1) = \cosh(x_2). Then, ex1+e−x12=ex2+e−x22\frac{e^{x_1} + e^{-x_1}}{2} = \frac{e^{x_2} + e^{-x_2}}{2}
step 4
This implies ex1+e−x1=ex2+e−x2e^{x_1} + e^{-x_1} = e^{x_2} + e^{-x_2}
step 5
Since the exponential function is one-to-one, x1=x2x_1 = x_2 or x1=−x2x_1 = -x_2
step 6
However, since cosh⁡(x)\cosh(x) is an even function, it is not injective because x1≠x2x_1 \neq x_2 can still satisfy cosh⁡(x1)=cosh⁡(x2)\cosh(x_1) = \cosh(x_2)
step 7
To determine the surjectivity, we need to check if for every y∈[1,∞)y \in [1, \infty), there exists an x∈Rx \in \mathbb{R} such that f(x)=yf(x) = y
step 8
Since cosh⁡(x)\cosh(x) ranges from 11 to ∞\infty as xx ranges over R\mathbb{R}, f(x)f(x) is surjective
step 9
Therefore, f(x)=cosh⁡(x)f(x) = \cosh(x) is surjective but not injective
B
Key Concept
Surjectivity and Injectivity of Hyperbolic Functions
Explanation
The function cosh⁡(x)\cosh(x) is surjective because it covers the entire range [1,∞)[1, \infty), but it is not injective because it is an even function, meaning it does not map distinct inputs to distinct outputs.
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