First slide

Introduction to definite integration

Question

$\int_{- 1}^{4} (2 x - 3) dx$ is equal to

Easy

A

$\frac{3}{2}$
B

$\frac{1}{3}$
C

$\frac{1}{2}$
D

0

Solution

Let $I = \int_{- 1}^{4} (2 x - 3) dx and f (x) = 2 x - 3$
then,
$\begin{matrix} \int f (x) dx = \int (2 x - 3) dx \\ = \frac{2 x^{2}}{2} - 3 x \\ = x^{2} - 3 x = F (x) (say) \end{matrix}$
Now, by the fundamental theorem of integral calculus, we get
$\begin{matrix} I = [F (x)]_{- 1}^{4} = F (4) - F (- 1) \\ = [(4)^{2} - 3 (4)] - [(- 1)^{2} - 3 (- 1)] \\ = (16 - 12) - (1 + 3) \\ = 4 - 4 = 0 \end{matrix}$

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