7/3
Evaluating the Definite Integral $\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx$
This problem requires us to evaluate a definite integral that includes the absolute value function, $|x|$. The function inside the integral is $f(x) = x|x|$. The absolute value function $|x|$ changes its definition at $x=0$. Specifically, $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$.
Because the interval of integration is from -1 to 2, which spans across $x=0$, we must split the definite integral into two parts based on the sign of $x$.
- For $x \ge 0$, $f(x) = x|x| = x(x) = x^2$. This applies to the interval $[0, 2]$.
- For $x < 0$, $f(x) = x|x| = x(-x) = -x^2$. This applies to the interval $[-1, 0)$.
Therefore, we can rewrite the given definite integral as the sum of two integrals:
\(\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx = \mathop \smallint \limits_{ - 1}^0 x\left| x \right|\;dx + \mathop \smallint \limits_{ 0}^2 x\left| x \right|\;dx\)
Now, substitute the correct form of $x|x|$ for each interval:
\(\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx = \mathop \smallint \limits_{ - 1}^0 \left( -x^2 \right)\;dx + \mathop \smallint \limits_{ 0}^2 \left( x^2 \right)\;dx\)
Calculating Each Part of the Definite Integral
Let's evaluate the first definite integral from -1 to 0:
\(\mathop \smallint \limits_{ - 1}^0 \left( -x^2 \right)\;dx\)
The antiderivative of $-x^2$ is \(-\frac{x^3}{3}\). Applying the limits of integration:
\(\left[ -\frac{x^3}{3} \right]_{-1}^0 = \left( -\frac{(0)^3}{3} \right) - \left( -\frac{(-1)^3}{3} \right)\)
\(= \left( 0 \right) - \left( -\frac{-1}{3} \right)\)
\(= 0 - \frac{1}{3}\)
\(= -\frac{1}{3}\)
Next, let's evaluate the second definite integral from 0 to 2:
\(\mathop \smallint \limits_{ 0}^2 \left( x^2 \right)\;dx\)
The antiderivative of $x^2$ is \(\frac{x^3}{3}\). Applying the limits of integration:
\(\left[ \frac{x^3}{3} \right]_{0}^2 = \left( \frac{(2)^3}{3} \right) - \left( \frac{(0)^3}{3} \right)\)
\(= \left( \frac{8}{3} \right) - \left( 0 \right)\)
\(= \frac{8}{3}\)
Finding the Total Value of the Definite Integral
To find the total value of the original definite integral, we add the results from the two parts:
\(\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx = \left( -\frac{1}{3} \right) + \left( \frac{8}{3} \right)\)
\(= \frac{-1 + 8}{3}\)
\(= \frac{7}{3}\)
The value of the definite integral $\mathop \smallint \limits_{ - 1}^2 x\left| x \right|\;dx$ is \(\frac{7}{3}\).
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
A parametric curve is defined \(x = cos\left(\frac{\Pi t}{2}\right) , Y= sin\left(\frac{\Pi t}{2}\right)\) in the range of \(0\leq t\leq 1\) . It is rotated about X-axis by 360°.
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if \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx=Ax+B\log |sin(x-a)|+ C\) where A, B and c are real constants then:
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