Question

Determine o valor das seguintes integrais definidas.
a) integral π/2 cos(θ)tg(θ)dθ
b) integral (3x^2 − x + 4)dx

Answer 1

a)

$\int\nolimits _{0}^{\frac{\pi }{2}}\cos( \theta )\tan( \theta ) d\theta =\int\nolimits _{0}^{\frac{\pi }{2}}\cancel{\cos( \theta )} \cdotp \frac{\sin( \theta )}{\cancel{\cos( \theta )}} d\theta \\ \int\nolimits _{0}^{\frac{\pi }{2}}\cos( \theta )\tan( \theta ) d\theta =-\cos( \theta )\Bigl|_{0}^{\pi /2} \\ \int\nolimits _{0}^{\frac{\pi }{2}}\cos( \theta )\tan( \theta ) d\theta =-\cos\left(\frac{\pi }{2}\right) -( -\cos( 0)) \\ \therefore\boxed{\boxed{\int\nolimits _{0}^{\frac{\pi }{2}}\cos( \theta )\tan( \theta ) d\theta =1}}$

b)

$\int _{0}^{1} 3x^{2} -x+4dx=\left[\cancel{3} \cdotp \frac{x^{3}}{\cancel{3}} -\frac{x^{2}}{2} +4x\right]_{0}^{1} \\ \int _{0}^{1} 3x^{2} -x+4dx=\left[ x^{3} -\frac{x^{2}}{2} +4x\right]_{0}^{1} \\ \int _{0}^{1} 3x^{2} -x+4dx=1^{3} -\frac{1^{2}}{2} +4\cdotp 1-\left( 0^{3} -\frac{0^{2}}{2} +4\cdotp 0\right) \\ \therefore\boxed{\boxed{\int _{0}^{1} 3x^{2} -x+4dx=\frac{9}{2}}}$