Question

The value of ∫(ax^3 + bx + c) dx , x ∈ [-2, 2] depends on

Answer 1

$\boxed{\underline{\textsf{Integration rule :}}}$

$\bold{\int x^{n}dx = \frac{1}{n+1}x^{n+1}+C}$

$\textsf{where C is integral constant.}$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\: \bold{I=\int (ax^{3}+bx+c)\:dx}$

$=\bold{\int ax^{3}\:dx+\int bx\:dx+\int c\:dx}$

$=\bold{a\int x^{3}\:dx + b\int x\:dx + c\int dx}$

$=\bold{a\frac{x^{3+1}}{3+1} + b\frac{x^{1+1}}{1+1} + cx}$

$=\bold{\frac{a}{4}x^{4} + \frac{b}{2}x^{2} + cx}$

$\textsf{Here, we have not considered integral}$
$\textsf{constant because it is a Particular}$
$\textsf{Integral.}$

$\textsf{Thus,}\:\bold{I_{-2}^{+2}}$

$=\bold{[\frac{a}{4}x^{4} + \frac{b}{2}x^{2} + cx]_{-2}^{+2}}$

$=\bold{[\frac{a}{4}(2)^{4} + \frac{b}{2}(2)^{2} + c(2)]}$

$\bold{-[\frac{a}{4}(-2)^{4} + \frac{b}{2}(-2)^{2} + c(-2)]}$

$=\bold{(\frac{16a^{4}}{4}+\frac{4b^{2}}{2}+2c)}$

$\bold{-(\frac{16a^{4}}{4}+\frac{4b^{2}}{2}-2c)}$

$=\bold{\frac{16a^{4}}{4}+\frac{4b^{2}}{2}+2c}$

$\bold{-\frac{16a^{4}}{4}-\frac{4b^{2}}{2}+2c}$

$=\bold{2c+2c}$

$=\bold{4c}$

$\to \boxed{\bold{\int_{-2}^{2} (ax^{3}+bx+c)\:dx=4c}}$

$\therefore \underline{\small{\textsf{the integral's value depends on c only.}}}$