What is the integral from-1 to+1 of x^{99}P_{100}(x) ?

The integral from-1 to+1 of x^{99}P_{100}(x) is (2P_{100})/(101)

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integral from-1 to+1 of x^{99}P_{100}(x)

\int_{- 1}^{+ 1} x^{99} P_{100} (x) d x

\frac{2 P _{100}}{101}

Solution steps

\int_{- 1}^{1} x^{99} P_{100} x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= P_{100} \cdot \int_{- 1}^{1} x x^{99} d x

Simplify

x x^{99} : x^{100}

= P_{100} \cdot \int_{- 1}^{1} x^{100} d x

Apply the Power Rule

: [\frac{x ^{101}}{101}]_{- 1}^{1}

= P_{100} [\frac{x ^{101}}{101}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{101}

= P_{100} \frac{2}{101}

Simplify

= \frac{2 P _{100}}{101}

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What is the integral from-1 to+1 of x^{99}P_{100}(x) ?
The integral from-1 to+1 of x^{99}P_{100}(x) is (2P_{100})/(101)