If int {({x^3} - 2{x^2} + 5){e^{3x}},dx} = e^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int (x^3 - 2x^2 + 5)e^{3x} dx$. Using the method of differentiation of the assumed form $e^{3x}(Ax^3 + Bx^2 + Cx + D)$,we have:$\frac{d}

Vedclass · Accepted Answer

$C + 2B = 0$

Vedclass · Answer

(C) Let $I = \int (x^3 - 2x^2 + 5)e^{3x} dx$. Using the method of differentiation of the assumed form $e^{3x}(Ax^3 + Bx^2 + Cx + D)$,we have:$\frac{d}{dx} [e^{3x}(Ax^3 + Bx^2 + Cx + D)] = e^{3x}(x^3 - 2x^2 + 5)$.Applying the product rule:$3e^{3x}(Ax^3 + Bx^2 + Cx + D) + e^{3x}(3Ax^2 + 2Bx + C) = e^{3x}(x^3 - 2x^2 + 5)$.Comparing coefficients of powers of $x$:$x^3: 3A = 1 \implies A = 1/3$.$x^2: 3B + 3A = -2 \implies 3B + 1 = -2 \implies 3B = -3 \implies B = -1$.$x^1: 3C + 2B = 0 \implies 3C - 2 = 0 \implies C = 2/3$.$x^0: 3D + C = 5 \implies 3D + 2/3 = 5 \implies 3D = 13/3 \implies D = 13/9$.Now check the options:$A) C + 3D = 2/3 + 13/3 = 15/3 = 5$ (Correct).$B) A + B + 2/3 = 1/3 - 1 + 2/3 = 0$ (Correct).$C) C + 2B = 2/3 + 2(-1) = 2/3 - 2 = -4/3 
eq 0$ (Incorrect).$D) A + B + C = 1/3 - 1 + 2/3 = 0$ (Correct).Thus,the incorrect statement is $C$.

If $\int {({x^3} - 2{x^2} + 5){e^{3x}}\,dx} = e^{3x} (Ax^3 + Bx^2 + Cx + D) + K$,then the statement which is incorrect is

Similar Questions

If $\int {\frac{{\cos 4x + 1}}{{\cot x - \tan x}}} dx = k\,\cos 4x + c$,then:

$\int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx = A \{23x + 2\ln |4\sin x + 5\cos x|\} + c$,then $A$ and $f(x)$ are

$\int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} = $

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \operatorname{Sin}^{-1} \sqrt{x}+c$,then $f(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module