Mathematics

$$-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c)  + d$$, then


SOLUTION
$$-2\int x^2e^{-2x}dx$$

$$=-2\left ( x^2\dfrac{-1}{2}e^{-2x}+\dfrac{1}{4}\left ( -2e^{-2x}x+e^{-2x} \right ) \right )+C$$

$$=e^{-2x}x^2+e^{-2x}x-\dfrac{1}{2}e^{-2x}+C$$

$$ -2\int x^2e^{-2x}dx=e^{-2x}x^2-e^{-2x}x-\dfrac{1}{2}e^{-2x}+C$$

$$\Rightarrow -2\int x^2e^{-2x}dx=e^{-2x}\left ( x^2+x-\dfrac{1}{2} \right )+C$$

$$\Rightarrow $$ $$e^{-2x}(ax^2+bx+c)+d=e^{-2x}\left ( x^2+x-\dfrac{1}{2} \right )+C$$

$$\Rightarrow a=1$$

$$\Rightarrow b=1$$

$$\Rightarrow c=\dfrac{-1}{2}$$

$$\Rightarrow d=C(constant)$$
View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
If $$ \int_{0}^{1} e^{x^{2}}(x-\alpha) d x=0, $$ then
  • A. $$

    \alpha<0

    $$
  • B. $$

    0<\alpha<1

    $$
  • C. $$

    \alpha=0

    $$
  • D. $$

    1<\alpha<2

    $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Integrate $$\int x{\sec ^2}x dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
If n is any positive integer, show that the integral part of $$ ( 3 + \sqrt7)^n $$ is an odd number. 

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate:-
$$\int_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2 - \operatorname{sinx} }}{{2 + \sin x}}} \right)dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate $$\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer