Passage

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts. 

$$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$$ $$\int\, u^{n}(x)v_{n}(x)\, dx$$ where $$v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $$\int P_{n}(x)\, Q(x)\, dx$$, where $$P_{n}(x)$$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.

Mathematics

If $$\int e^{2x}. x^{4}\, dx\, =\, \displaystyle \frac{e^{2x}}{2} f(x)\, +\, C$$ then f(x) is equal to


ANSWER

$$x^{4}\, -\, 2x^{3}\, +\, 3x^{2}\, -\, 3x\, +\, \displaystyle \frac{3}{2}$$


SOLUTION
$$I=\int e^{2x}. x^{4} dx$$
Applying integration by parts,
$$\displaystyle = x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -\int { { 4x }^{ 3 } } \frac { e^{ 2x } }{ 2 } dx$$

$$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -2\int { { x }^{ 3 } } e^{ 2x }dx$$
Again applying integration by parts
$$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -2{ x }^{ 3 }\frac { e^{ 2x } }{ 2 } +2\int { { 3x }^{ 2 } } \frac { e^{ 2x } }{ 2 } dx$$

$$\Rightarrow \displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+3 \int { { x }^{ 2 } } e^{ 2x }dx$$

Again applying integration by parts, we get
$$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+3x^{ 2 }\, \frac { e^{ 2x } }{ 2 } -3\int { { 2x } } \frac { e^{ 2x } }{ 2 } dx$$

$$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-3 \int { { x } } e^{ 2x }dx$$

Again applying integration by parts, we get
$$\displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-\frac { 3 }{ 2 } xe^{ 2x }+3 \int { \frac { e^{ 2x } }{ 2 }  } dx$$

$$\Rightarrow \displaystyle I=x^{ 4 }\, \frac { e^{ 2x } }{ 2 } -{ x }^{ 3 }e^{ 2x }+\frac { 3 }{ 2 } x^{ 2 }\, e^{ 2x }-\frac { 3 }{ 2 } xe^{ 2x }+\frac { 3 }{ 4 } e^{ 2x }+C$$

$$\Rightarrow \displaystyle I=\dfrac{e^{2x}}{2} (x^4-2x^3+3x^2-3x+\frac{3}{2})+C$$

On comparing with given , we get
$$f(x)=x^4-2x^3+3x^2-3x+\dfrac{3}{2}$$
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Single Correct Hard Published on 17th 09, 2020
Mathematics

If $$\int (x^{3}\, -\, 2x^{2}\, +\, 3x\, -\, 1)\, cos2x\, dx\, =\, \displaystyle \frac{sin 2x}{4}u(x)\, +\, \frac{cos 2x}{8}v(x)\, +\, c$$, then


ANSWER

$$u(x)\, =\, 2x^{3}\, -\, 4x^{2}\, +\, 3x$$


SOLUTION
$$\int { \left( { x }^{ 2 }-2{ x }^{ 2 }+3x-1 \right)  } \cos { 2x } dx\\ =\int { { x }^{ 2 } } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x }  } dx+\int { 3x\cos { 2x }  } dx-\int { \cos { 2x }  } dx\\ =\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } -\frac { 3 }{ 2 } \int { { x }^{ 2 } } \sin { 2x } dx-2\int { { x }^{ 2 }\cos { 2x }  } dx+3\int { x } \cos { 2x } d-\int { \cos { 2x } dx } $$
$$\displaystyle =\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } +\frac { 1 }{ 2 } { x }^{ 3 }\sin { 2x } +\frac { 3 }{ 2 } \int { x } \cos { 2x } dx-2\int { { x }^{ 2 }\cos { 2x } dx } -\int { \cos { 2x }  } dx$$
$$\displaystyle ={ x }^{ 3 }\sin { x } \cos { x- } { x }^{ 2 }\sin { 2x } +\frac { 3 }{ 4 } { x }^{ 2 }\cos { 2x } \frac { 3 }{ 4 } x\sin { 2x } -x\cos { 2x } +\frac { 3 }{ 8 } \cos { 2x } $$
$$\displaystyle =\frac { 1 }{ 4 } \left( { 2x }^{ 3 }-{ 4x }^{ 2 }+3x \right) \sin { 2x } +\frac { 1 }{ 8 } \left( { 12x }^{ 3 }-{ 16x }^{ 2 }+6x \right) \cos { 2x } $$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
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