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Question 32

The value of

\int_1^2 (4x^2 - x + 5) dx

is ....

$\frac{28}{3}$

$\frac{77}{3}$

$\frac{77}{6}$

$6$

$7$

To solve the definite integral $\int_1^2 (4x^2 - x + 5) dx$ , we integrate first

\int_1^2 (4x^2 - x + 5) dx = \left[\frac{4}{3}x^3 - \frac{1}{2}x^2 + 5x\right]_1^2

Substitute $x = 2$ and $x = 1$

= \left(\frac{4}{3}(2)^3 - \frac{1}{2}(2)^2 + 5(2)\right) - \left(\frac{4}{3}(1)^3 - \frac{1}{2}(1)^2 + 5(1)\right)

= \left(\frac{4}{3} \cdot 8 - \frac{1}{2} \cdot 4 + 10\right) - \left(\frac{4}{3} - \frac{1}{2} + 5\right)

= \left(\frac{32}{3} - 2 + 10\right) - \left(\frac{4}{3} - \frac{1}{2} + 5\right)

= \left(\frac{32}{3} + 8\right) - \left(\frac{4}{3} - \frac{1}{2} + 5\right)

= \frac{32}{3} + 8 - \frac{4}{3} + \frac{1}{2} - 5

= \frac{32 - 4}{3} + 8 - 5 + \frac{1}{2}

= \frac{28}{3} + 3 + \frac{1}{2}

= \frac{56}{6} + \frac{18}{6} + \frac{3}{6}

= \frac{56 + 18 + 3}{6} = \frac{77}{6}

Therefore, the value of $\int_1^2 (4x^2 - x + 5) dx = \frac{77}{6}$

Exercises