First slide

Introduction to definite integration

Question

Let $f : R \to R$ be a function defined by $f (x) = \{\begin{cases} [x], x \leq 2 \\ 0, x > 2 \end{cases}$ where $[x]$ is the greatest integer less than or equal to $x$ . if $I = \int_{- 1}^{2} \frac{x f (x^{2})}{2 + f (x + 1)} d x$
then the value of $(4 I - 1)$ is $—$

Moderate

Solution

$\begin{array}{l} I = \int_{- 1}^{2} \frac{x [x^{2}]}{2 + [x + 1]} d x = \int_{- 1}^{2} \frac{x [x^{2}]}{3 + [x]} d x \\ = \int_{- 1}^{0} \frac{0}{3 - 1} d x + \int_{0}^{1} \frac{0}{3 + 1} d x + \int_{1}^{\sqrt{2}} \frac{x \cdot 1}{3 + 1} d x + 0 \\ = \frac{1}{4} {[\frac{x^{2}}{2}]}_{1}^{\sqrt{2}} = \frac{2 - 1}{8} = \frac{1}{4} \\ ∴ 4 I - 1 = 0 \end{array}$

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