Ejercicios de exámenes 2

\[\displaystyle\int_1^4 \left( \sqrt{x} + f(x)\right)dx=5\] \[\displaystyle\int_1^4 x^{\frac{1}{2}}dx+ \int_1^4 f(x) dx=5\] \[\displaystyle\int_1^4 \dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}dx + \int_1^4 f(x) dx=5\] \[\left. \dfrac{2x^{\frac{3}{2}}}{3} \right |_1^4 +\int_1^4 f(x) dx=5\] \[\left[\left(\dfrac{2(4)^{\frac{3}{2}}}{3}\right)-\left(\dfrac{2(1)^{\frac{3}{2}}}{3}\right)\right] +\int_1^4 f(x) dx=5\] \[\left(\dfrac{16}{3}-\dfrac{2}{3}\right)+\int_1^4 f(x) dx=5\] \[\dfrac{14}{3}+\int_1^4 f(x) dx=5\] \[\int_1^4 f(x) dx=5-\dfrac{14}{3}\] \[\int_1^4 f(x) dx=\dfrac{1}{3}\]

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Datos:

• \(h(x)=(x^3-6x)f(x)\)
• \(f(1)=5\)
• \(f'(1)=9\)

Hallar \(h'(1)\) (la derivada de una multiplicación):
\[h'(x)= (x^3-6x)'f(x)+ (x^3-6x)f'(x)\] \[h'(x)=(3x^2-6)f(x)+(x^3-6x)f'(x)\] \[h'(1)=(3(1)^2-6)f(1)+((1)^3-6(1))f'(1)\] \[h'(1)=\left\lbrace [3(1)^2-6]5\right\rbrace +\left\lbrace[(1)^3-6(1)]9 \right\rbrace\] \[h'(1)=[(-3)5]+[(-5)9]\] \[h'(1)=(-15)+(-45)\] \[h'(1)=-60\]

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Hallar \(\int_{-1}^{1} f(x)dx=\) \[\int_{-1}^{1} \left[3f(x)-2 \right]dx=4\] \[3\int_{-1}^{1} f(x) dx -2\int_{-1}^{1} dx=4\] \[3\int_{-1}^{1} f(x) dx - \left. 2x \right|_{-1}^{1}=4\] \[3\int_{-1}^{1} f(x) dx -\left[ (2(1))-(2(-1) \right]=4\] \[3\int_{-1}^{1} f(x) dx -4 =4\] \[3\int_{-1}^{1} f(x) dx=4+4\] \[\int_{-1}^{1} f(x) dx=\dfrac{8}{3}\]

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Hallar los ceros de \(f\) en \([0;2\pi]\) \[f(x)=0\] \[2\cos^2 (x)+\cos (x)=0\] \[\cos (x)(2\cos (x) +1)=0\] \[\Rightarrow\, \cos (x) = 0 \;\vee\, 2\cos(x)+1=0\] \[\Rightarrow\, \cos (x) = 0 \;\vee\, \cos(x)=-\dfrac{1}{2}\]

• \(\cos(x)=0\), período de los ceros de \(\cos\) es \(k\pi\) \[x=\dfrac{\pi}{2}+k\pi\quad (k \in \mathbb{Z})\]

  • \(k=0\) \[x_1=\dfrac{\pi}{2}+(0)\pi= \dfrac{\pi}{2}\]

  • \(k=1\) \[x_2=\dfrac{\pi}{2}+(1)\pi=\dfrac{3\pi}{2}\]

• \(\cos(x)=-\dfrac{1}{2}\), en \(120^o\) y \(240^o\). \[x_3=\dfrac{2\pi}{3} \, \wedge \, x_4=\dfrac{4\pi}{3}\]

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¿Para qué valor de k se cumple que el conjunto de negatividad de la función \(f(x)=2^{-k+x}-4\) es igual a \((-\infty;3)\)?

• En \(x=3\), \(f(x)=0\)

\[f(x)<0\] \[f(3)=0\] \[2^{-k+3}-4=0\] \[2^{-k+3}=4\] \[\log_2 2^{-k+3}=\log_2 4\] \[-k+3=2\] \[-k=2-3\] \[-k(-1)=-1(-1)\] \[k=1\]