Question

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

• A. $C+3D=5$
• B. $A+B+\frac{2}{3}=0$
• C. $C+2B=0$
• D. $A+B+C=0$

Answer 1

Answer: (C)

$C+2B=0$

There's a very quick way to do sums like these of the form;

$\int e^{x}p\left(x\right)dx$

Where $p\left(x\right)$ is a polynomial in x.

We know that;

$\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f\left(x\right)+C$

Let $f\left(x\right)+f^{\prime }\left(x\right)=p\left(x\right)$

Therefore;

$f\left(x\right)=p\left(x\right)-f^{\prime }\left(x\right)...eqn1f^{\prime }\left(x\right)=p^{\prime }\left(x\right)-f^{\prime \prime }\left(x\right)f^{\prime \prime }\left(x\right)=p^{\prime \prime }\left(x\right)-f^{\prime \prime \prime }\left(x\right)...$

Let's now substitute the values of $f^{\prime },f^{\prime \prime },f^{\prime \prime \prime },f^{\prime \prime \prime \prime },....$ to $eqn1$

We get;

$f\left(x\right)=p\left(x\right)-p^{\prime }\left(x\right)+p^{\prime \prime }\left(x\right)-p^{\prime \prime \prime }\left(x\right)....$

Hence the final formula is;

$\int e^{x}p\left(x\right)dx=e^x(p\left(x\right)-p^{\prime }\left(x\right)+p^{\prime \prime }\left(x\right)-p^{\prime \prime \prime }\left(x\right)...)+C$

Although this formula can be used for any function, polynomial functions work best;

---------------------------------------

substitute $3x$ as $t$

$3dx=dtI=\int [{\left(\frac{t}{3}\right)}^3-2{\left(\frac{t}{3}\right)}^2+5]e^t\frac{dt}{3}I=\frac{1}{3^4}\int (t^3-6t^2+135)e^{t}dt$

We can now see the above form is like $\int e^{x}p\left(x\right)dx$

$\therefore$

let, $p\left(t\right)=t^3-6t^2+135p^{\prime }\left(t\right)=3t^2-12t{p}^{\prime \prime }\left(t\right)=6t-12p^{\prime \prime \prime }\left(t\right)=6p""\left(t\right)=0...\therefore p\left(t\right)-p^{\prime }\left(t\right)+p^{\prime \prime }\left(t\right)-p^{\prime \prime \prime }\left(t\right)...=(t^3-6t^2+135)-(3t^2-12t)+(6t-12)-6+0-0+\mathrm{0...}=t^3-9t^2+18t+117I=\frac{1}{3^4}[t^3-9t^2+18t+117]e^t+kI=\frac{1}{3^4}[{\left(3x\right)}^3-9{\left(3x\right)}^2+18\left(3x\right)+117]e^{3x}+kI=[\left(\frac{1}{3}\right)x^3+(-1)x^2+\left(\frac{2}{3}\right)x+\left(\frac{13}{9}\right)]e^{3x}+k$

$\therefore A=\frac{1}{3}B=-1C=\frac{2}{3}D=\frac{13}{9}$

Hence the only incorrect option is C.