Question
Download Solution PDF\(\rm \int_{-1 }^{1} \left | x \right |dx\) का मान ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
- \(\rm \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\)
गणना:
दिया गया, \(\rm \int_{-1 }^{1} \left | x \right |dx\)
जैसा कि हम जानते हैं कि, \(f\left( x \right) = \left| x \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;\;x < 0}\\ {x,\;\;x \ge 0} \end{array}} \right.\)
तो f(x) = - x जब -1 < x < 0 और f(x) = x जब 0 < x < 1
जैसा कि हम जानते हैं कि, \(\rm \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\)
\(\rm \Rightarrow\int_{-1 }^{1} \left | x \right |dx=\int_{-1 }^{0} (-x)dx+\int_{0 }^{1} (x)dx\)
\(\rm \Rightarrow\int_{-1 }^{0} (-x)dx+\int_{0 }^{1} (x)dx=\left [ \frac{-x^{2}}{2} \right ]_{-1}^{0}+\left [ \frac{x^{2}}{2} \right ]_{0}^{1}=\frac{1}{2}+\frac{1}{2}=1\)
इसलिए, सही विकल्प 4 है।
Last updated on Jun 30, 2025
->Indian Airforce Agniveer (02/2026) Notification has been released. Interested candidates can apply between 11th July to 31st July 2025.
->The Examination will be held 25th September 2025 onwards.
-> Earlier, Indian Airforce Agniveer Group X 2025 Last date had been extended.
-> Candidates applied online from 7th to 2nd February 2025.
-> The online examination was conducted from 22nd March 2025 onwards.
-> The selection of the candidates will depend on three stages which are Phase 1 (Online Written Test), Phase 2 ( DV, Physical Fitness Test, Adaptability Test), and Phase 3 (Medical Examination).
-> The candidates who will qualify all the stages of selection process will be selected for the Air Force Group X posts & will receive a salary ranging of Rs. 30,000.
-> This is one of the most sought jobs. Candidates can also check the Airforce Group X Eligibility here.