3 ∫ 1 ( X 2 + 3 X ) D X - Mathematics

प्रश्न

\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]

बेरीज

उत्तर

\[\text{Here }a = 1, b = 3, f\left( x \right) = x^2 + 3x, h = \frac{3 - 1}{n} = \frac{2}{n}\]

Therefore,

\[ \int_1^3 \left( x^2 + 3x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ 1 + 3 + \left( 1 + h \right)^2 + 3\left( 1 + h \right) + \left( 1 + 2h \right)^2 + 3\left( 1 + 2h \right) + . . . . . . . . . + \left( \left( n - 1 \right)h \right)^2 + 3\left( \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ n + h^2 \left( 1^2 + 2^2 + . . . . . . . . . . . . . . \left( n - 1 \right)^2 \right) + 2h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) + 3n + 3h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) \right]\]

\[ = \lim_{h \to 0} h\left[ 4n + h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 5h\frac{n\left( n - 1 \right)}{2} \right]\]

\[ = \lim_{n \to 0 } \left[ 8 + \frac{4}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) + 10\left( 1 - \frac{1}{n} \right) \right]\]

\[ = 8 + \frac{8}{3} + 10\]

\[ = \frac{62}{3}\]

shaalaa.com

Definite Integrals

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 67 Q 66 Q 68

पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12

पाठ 20 Definite Integrals
Revision Exercise | Q 67 | पृष्ठ १२३

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/2} \cos^3 x\ dx\]

\[\int\limits_0^{\pi/2} x \cos\ x\ dx\]

\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]

\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]

\[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

If f(x) is a continuous function defined on [−a, a], then prove that

\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]

\[\int\limits_0^3 \left( x + 4 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]