Question

If ${\int }_a^b{x}^{3}dx=0$ and if ${\int }_a^b{x}^{2}dx=\frac{2}{3}$, find a and b respectively.

• A. $\frac{1}{4}$, 1
• B. -1, 4
• C. -1, 1
• D. 2, 4

Answer 1

The correct option is C -1, 1

${\int }_a^b{x}^{3}dx=0$

$[\frac{x^4}{4}{]}_a^b=0$

$\frac{1}{4}(b^4-a^4)=0$

$b^4-a^4=0$

$(b^2-a^2)(b^2+a^2)=0$

$b^2-a^2=0$

$b=-a$

Now, ${\int }_a^b{x}^{2}dx=\frac{2}{3}$

$(\frac{x^3}{3}{)}_a^b=\frac{2}{3}$

$\frac{1}{3}(b^3-a^3)=\frac{2}{3}$

$b^3-a^3=2$

$(-a{)}^3-a^3=2$

$-2a^3=2$

$a^3=-1$

$a=-1$

Thus, $a=-1,b=1$