Q.Evaluate $\displaystyle\int_1^2(3x^2-2x)\,dx$.

$\Gamma(n)=(n-1)!$ for a positive integer $n$, hence $\Gamma(10)=9!$.

The Gamma function is defined by the definite integral $\Gamma(n)=\int_0^{\infty} e^{-x}x^{\,n-1}\,dx$, and for a positive integer $n$ it reduces to a factorial:

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$\int \cos x\,dx = \sin x$, and evaluating between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$ gives $2$.

Using the Fundamental Theorem of Calculus with the antiderivative $\int\cos x\,dx=\sin x$:

$\int_{-\pi/2}^{\pi/2}\cos x\,dx = \big[\sin x\big]_{-\pi/2}^{\pi/2} = \sin\tfrac{\pi}{2} - \sin\!\left(-\tfrac{\pi}{2}\right).$

Since $\sin\tfrac{\pi}{2}=1$ and $\sin(-\tfrac{\pi}{2})=-1$:

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Integrate $x^{1/2}$ and $x^{-1/2}$ by the power rule, then evaluate from $0$ to $4$ to get $\dfrac{28}{3}$, option (a).

Antiderivative.

$\int\Big(x^{1/2}+x^{-1/2}\Big)dx=\frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}=\frac{2}{3}x^{3/2}+2\sqrt{x}.$

Apply the limits. …

$\Gamma(3/2)=\tfrac12\Gamma(1/2)=\tfrac{\sqrt{\pi}}{2}$.

The Gamma function satisfies the recurrence $\Gamma(n+1)=n\,\Gamma(n)$, and $\Gamma\left(\tfrac12\right)=\sqrt{\pi}$. Writing $\tfrac32=\tfrac12+1$,

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$\int_0^\infty e^{-2x}\,dx = \tfrac12$.

Use the standard result $\int e^{ax}\,dx = \dfrac{e^{ax}}{a}$ with $a = -2$:

$\int_0^\infty e^{-2x}\,dx = \left[\frac{e^{-2x}}{-2}\right]_0^\infty = \left[-\frac{1}{2}e^{-2x}\right]_0^\infty.$

Apply the limits. As $x \to \infty$, $e^{-2x} \to 0$; at $x = 0$, $e^{0} = 1$:

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Apply the Gamma-function recurrence $\Gamma(n+1) = n\,\Gamma(n)$ twice to get $\Gamma(n+2) = (n+1)n\,\Gamma(n)$, cancel $\Gamma(n)$, and solve $(n+1)n = 90$.

Step 1 — Reduce $\Gamma(n+2)$. Using $\Gamma(m+1) = m\,\Gamma(m)$:

$\Gamma(n+2) = (n+1)\,\Gamma(n+1) = (n+1)\,n\,\Gamma(n).$

Step 2 — Substitute into the given relation.

$(n+1)\,n\,\Gamma(n) = 90\,\Gamma(n).$ …