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Integrate sqrt(x^3 + 1) from 0 to 2
\int_{0}^{2} \sqrt{x^{3} + 1}\, dx
चरण दर चरण
- \int_{0}^{2} \sqrt{x^{3} + 1}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- \int \sqrt{x^{3} + 1}\, dx = \frac{x \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {x^{3} e^{i \pi}} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}
This one needs a special technique; the computer algebra system gives the antiderivative directly.
- F(2) - F(0) = \left(2 {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {8 e^{i \pi}} \right)}\right) - \left(0\right)
Fundamental theorem of calculus: plug in the limits.
- = 2 {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {8 e^{i \pi}} \right)}
Simplify.
जवाब दिखाएँ
2 {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {8 e^{i \pi}} \right)}