Question

Calculo1-Integral help +++

Answer 1

$cosh(x)= \frac{1}{2}(e^{-x}- e^x)$

$\frac{1}{2} \int\limits^1_0 {e^{-x}+e^x} \, dx \\\\\ \frac{1}{2}\left( \int\limits^1_0 e^{-x}}dx+ \int\limits^1_0 {e^{x}}dx\right) \\\\\\ = \frac{1}{2}\left(\left -e^{-x}\right |^1_0 +\left e^{x}\right |^1_0 \right)\\\\= \frac{1}{2}\left [(-e^{-1}+1) +(e^1-1)\right]\\\\ \frac{1}{2} (- e^{-1} +e )$

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B)$4* \int\limits^{ \frac{1}{ \sqrt{2} } }_{ \frac{1}{2} } { \frac{1}{ \sqrt{1-x^2} } } \, dx$

fazendo uma substituição trigonométrica

$\bmatrix x=1*sen(\theta)\\\\ dx= cos(\theta)d\theta\\\\1-x^2 = 1-sen^2(\theta) = cos^2(\theta)\end$

ficando

$4* \int\limits^{ \frac{1}{ \sqrt{2} } }_{ \frac{1}{2} } \frac{1}{ \sqrt{cos^2(\theta)} }*cos(\theta)d\theta \\\\ =4\int\limits^{ \frac{1}{ \sqrt{2} } }_{ \frac{1}{2} } d\theta\\\\ = 4*\left[ \theta \right]^{ \frac{1}{ \sqrt{2} } }_{ \frac{1}{2} }\\\\ 4*\left[arcsen(x) \right]^{ \frac{1}{ \sqrt{2} } }_{ \frac{1}{2} }\\\\ = 4*\left[arcsen( \frac{1}{ \sqrt{2}}) - arcsen( \frac{1}{2} ) \right ]$

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$\int\limits^1_{-1} {e^{u+1}} \, du$

substituição
$x= u+1\\\\ \frac{dx}{du}=1 \\\\dx=du$

$\int\limits^1_{-1} {e^x} \, dx = e^x\left|^1_{-1} = e^{u+1}\left|_{-1}^1 = e^2 - e^0 = e^2-1$