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Calcule \[\int _{0}^3 3x-4\; dx.\]
xmhintEste é um problema que usa o teorema fundamental do cálculo. O teorema
fundamental do cálculo diz que, se \(F(x) = \int f(x)\; dx\) é uma antiderivada qualquer de \(f(x)\), então
\[\int _a^b f(x)\; dx = F(b)-F(a).\]
xmhintAplicando isso ao problema em questão, vemos que \[\int 3x-4\; dx = \frac {3}{2}x^2 -4x +C, \] de modo que podemos
escolher \(F(x) = \frac {3}{2}x^2-4x\) com \(C=0\). Então, \begin{align*} \int _0^3 3x-4\; dx &= \left .\frac {3}{2}x^2-4x+C\right |_0^3 \\ &= F(3)-F(0) \\ &= \left (\frac {3}{2}3^2-4(3)\right )-\left (\frac {3}{2}(0^2)-4(0)\right ) \\ &= \frac {27}{2}-12 \\ &= \frac {3}{2} \end{align*}
\(-\frac {3}{2}\) \(-\frac {51}{2}\) \(-15\) \(\frac {3}{2}\) \(15\)