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DivGradCurl
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I need some help with the following problems. Any help is highly appreciated.
1. If f is continuous on \mathbb{R}, prove that
\int _a ^b f(-x) \: dx = \int _{-b} ^{-a} f(x) \: dx
For the case where f(x) \geq 0 and 0 < a < b, draw a diagram to interpret this equation geometrially as an equality of areas.
2. If f is continuous on \mathbb{R}, prove that
\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(x) \: dx
For the case where f(x) \geq 0, draw a diagram to interpret this equation geometrially as an equality of areas.
3. If a and b are positive numbers, show that
\int _0 ^1 x^a (1 - x) ^b \: dx = \int _0 ^1 x^b (1 - x) ^a \: dx
Here is what I've got so far:
1. Consider the left-hand side
\int _a ^b f(-x) \: dx
and apply the substitution rule:
u=-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = - du
u(b)=-b
u(a)=-a
\int _a ^b f(-x) \: dx = -\int _{-a} ^{-b} f(u) \: du = \int _{-b} ^{-a} f(u) \: du = \int _{-b} ^{-a} f(x) \: dx
2. Consider the left-hand side
\int _a ^b f(x + c) \: dx
and apply the substitution rule:
u=x+c \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx = du
u(b)=b+c
u(a)=a+c
\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(u) \: du = \int _{a+c} ^{b+c} f(x) \: dx
3. Consider the left-hand side
\int _0 ^1 x^a (1 - x) ^b \: dx
and apply the substitution rule:
u=1-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = -du
u(1)=0
u(0)=1
\int _0 ^1 x^a (1 - x) ^b \: dx = \int _1 ^0 u^b (1 - u) ^a \: du = \int _0 ^1 x^b (1 - x) ^a \: dx
1. If f is continuous on \mathbb{R}, prove that
\int _a ^b f(-x) \: dx = \int _{-b} ^{-a} f(x) \: dx
For the case where f(x) \geq 0 and 0 < a < b, draw a diagram to interpret this equation geometrially as an equality of areas.
2. If f is continuous on \mathbb{R}, prove that
\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(x) \: dx
For the case where f(x) \geq 0, draw a diagram to interpret this equation geometrially as an equality of areas.
3. If a and b are positive numbers, show that
\int _0 ^1 x^a (1 - x) ^b \: dx = \int _0 ^1 x^b (1 - x) ^a \: dx
Here is what I've got so far:
1. Consider the left-hand side
\int _a ^b f(-x) \: dx
and apply the substitution rule:
u=-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = - du
u(b)=-b
u(a)=-a
\int _a ^b f(-x) \: dx = -\int _{-a} ^{-b} f(u) \: du = \int _{-b} ^{-a} f(u) \: du = \int _{-b} ^{-a} f(x) \: dx
2. Consider the left-hand side
\int _a ^b f(x + c) \: dx
and apply the substitution rule:
u=x+c \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx = du
u(b)=b+c
u(a)=a+c
\int _a ^b f(x + c) \: dx = \int _{a+c} ^{b+c} f(u) \: du = \int _{a+c} ^{b+c} f(x) \: dx
3. Consider the left-hand side
\int _0 ^1 x^a (1 - x) ^b \: dx
and apply the substitution rule:
u=1-x \Rightarrow \frac{du}{dx} = -1 \Rightarrow dx = -du
u(1)=0
u(0)=1
\int _0 ^1 x^a (1 - x) ^b \: dx = \int _1 ^0 u^b (1 - u) ^a \: du = \int _0 ^1 x^b (1 - x) ^a \: dx
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