What is the integral from 0 to sqrt(x of)t^3 ?

The integral from 0 to sqrt(x of)t^3 is (x^2)/4

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integral from 0 to sqrt(x of)t^3

\int_{0}^{x} t^{3} d t

\frac{x ^{2}}{4}

Solution steps

\int_{0}^{x} t^{3} d t

Apply the Power Rule

: [\frac{t ^{4}}{4}]_{0}^{x}

= [\frac{t ^{4}}{4}]_{0}^{x}

Compute the boundaries:

\frac{x ^{2}}{4}

= \frac{x ^{2}}{4}

\int_{0}^{1} 6 (\frac{y}{x})^{4} d y

\int_{1}^{25} \frac{1}{1 + x} d x

\int_{1}^{\infty} \frac{1}{n ^{3}}

\int_{1}^{\infty} \frac{1}{n ^{4}}

\int_{- 2}^{5} (10 x - x^{2}) d x

What is the integral from 0 to sqrt(x of)t^3 ?
The integral from 0 to sqrt(x of)t^3 is (x^2)/4