5 ∫ 0 4 √ X + 4 4 √ X + 4 + 4 √ 9 − X D X - Mathematics

प्रश्न

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

योग

उत्तर

\[Let I = \int_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x......................(1)\]
\[I = \int_0^5 \frac{\sqrt[4]{9 - x}}{\sqrt[4]{9 - x} - \sqrt[4]{x + 4}}dx ...........................\left[\text{Using }\int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[I = - \int_0^5 \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx ...................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} - \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx \]
\[ = \int_0^5 \frac{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx\]
\[ = \int_0^5 dx\]
\[ = \left[ x \right]_0^5 \]
\[ = 5\]
\[Hence\ I = \frac{5}{2}\]

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Definite Integrals

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 13 Q 12 Q 14

अध्याय 20: Definite Integrals - Exercise 20.4 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12

अध्याय 20 Definite Integrals
Exercise 20.4 | Q 13 | पृष्ठ ६१

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