Question Number 209659 by mokys last updated on 17/Jul/24
Answered by efronzo1 last updated on 18/Jul/24
$$\:\:\:\mathrm{S}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{2}^{\mathrm{x}} \:\mathrm{dx}\:+\:\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\left(\mathrm{3}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$
Answered by mr W last updated on 18/Jul/24
Commented by mr W last updated on 18/Jul/24
![T=∫_0 ^1 (3−x^2 −2^x )dx =[3x−(x^3 /3)−(2^x /(ln 2))]_0 ^1 =3−(1/3)−((2−1)/(ln 2))=(8/3)−(1/(ln 2)) S+T=(2/3)×3×(√3)=2(√3) ⇒S=2(√3)−(8/3)+(1/(ln 2))≈2.24](https://www.tinkutara.com/question/Q209693.png)
$${T}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}−{x}^{\mathrm{2}} −\mathrm{2}^{{x}} \right){dx} \\ $$$$\:\:\:=\left[\mathrm{3}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{2}−\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}} \\ $$$${S}+{T}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{3}×\sqrt{\mathrm{3}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{S}=\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{2}.\mathrm{24} \\ $$
Answered by lepuissantcedricjunior last updated on 18/Jul/24
![f(x)=3−x^2 ;g(x)=2^x 2^x +x^2 −3=0=>x=1=>0≤x≤1 S=∫_0 ^1 ∫_0 ^2^x dx=[(1/(ln2))2^x ]_0 ^1 =(1/(ln2))≈(1/(0.69))≈((100)/(69))](https://www.tinkutara.com/question/Q209689.png)
$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{3}−\boldsymbol{{x}}^{\mathrm{2}} ;\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\mathrm{2}^{\boldsymbol{{x}}} \\ $$$$\mathrm{2}^{\boldsymbol{{x}}} +\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}=>\boldsymbol{{x}}=\mathrm{1}=>\mathrm{0}\leqslant\boldsymbol{{x}}\leqslant\mathrm{1} \\ $$$$\boldsymbol{{S}}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{2}^{\boldsymbol{{x}}} } \boldsymbol{{dx}}=\left[\frac{\mathrm{1}}{\boldsymbol{{ln}}\mathrm{2}}\mathrm{2}^{\boldsymbol{{x}}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\boldsymbol{{ln}}\mathrm{2}}\approx\frac{\mathrm{1}}{\mathrm{0}.\mathrm{69}}\approx\frac{\mathrm{100}}{\mathrm{69}} \\ $$