Question Number 199351 by Tawa11 last updated on 01/Nov/23
Answered by MM42 last updated on 02/Nov/23
![a+b=2 ∫_0 ^4 (a(√x)+bx)dx=((2a)/3)x(√x)+(b/2)x^2 ]_0 ^4 =8 ⇒((16)/3)a+8b=8−8a ⇒a=3 & b=−1 ✓ (c),(d)](Q199354.png)
$${a}+{b}=\mathrm{2} \\ $$$$\left.\int_{\mathrm{0}} ^{\mathrm{4}} \left({a}\sqrt{{x}}+{bx}\right){dx}=\frac{\mathrm{2}{a}}{\mathrm{3}}{x}\sqrt{{x}}+\frac{{b}}{\mathrm{2}}{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{4}} \:=\mathrm{8} \\ $$$$\Rightarrow\frac{\mathrm{16}}{\mathrm{3}}{a}+\mathrm{8}{b}=\mathrm{8}−\mathrm{8}{a} \\ $$$$\Rightarrow{a}=\mathrm{3}\:\:\:\&\:\:\:{b}=−\mathrm{1}\:\:\checkmark\:\left({c}\right),\left({d}\right) \\ $$$$ \\ $$
Commented by Tawa11 last updated on 19/Nov/23
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$