Advertisements
Advertisements
प्रश्न
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
बेरीज
Advertisements
उत्तर
\[f'\left( x \right) = 8 x^3 - 2x f\left( 2 \right) = 8\]
\[f'\left( x \right) = 8 x^3 - 2x\]
\[\int{f}'\left( x \right)dx = \int\left( 8 x^3 - 2x \right)dx\]
\[ = 8\int x^3 dx - 2\ ∫ \text{ x dx}\]
\[f\left( x \right) = 8 \left[ \frac{x^4}{4} \right] - 2 \times \frac{x^2}{2} + C\]
\[f\left( x \right) = 2 x^4 - x^2 + C\]
\[f\left( 2 \right) = 8 \left( Given \right)\]
\[f\left( 2 \right) = 2 \times 2^4 - 2^2 + C\]
\[8 = 32 - 4 + C\]
\[C = - 20\]
\[ \therefore f\left( x \right) = 2 x^4 - x^2 - 20\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int \cot^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int x e^x \text{ dx }\]
\[\int x^3 \cos x^2 dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int {cosec}^3 x\ dx\]
\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{ dx }\]
\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int \sin^4 2x\ dx\]
\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int \sin^5 x\ dx\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
