What is the integral from 0 to 3 of t/(sqrt(t^2+16)) ?

The integral from 0 to 3 of t/(sqrt(t^2+16)) is 1

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integral from 0 to 3 of t/(sqrt(t^2+16))

\int_{0}^{3} \frac{t}{t ^{2} + 16} d t

1

Solution steps

\int_{0}^{3} \frac{t}{t ^{2} + 16} d t

Apply u-substitution

: \int_{16}^{25} \frac{1}{2 u} d u

= \int_{16}^{25} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{16}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{25}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{16}^{25}

Apply radical rule:

= \frac{1}{2} [2 u]_{16}^{25}

Compute the boundaries:

2

= \frac{1}{2} \cdot 2

Simplify

= 1

\int_{3}^{4} 2 x d x

\int_{- 1}^{1} (x^{3}) d x

\int_{- 1}^{2} (x^{3} + x) d x

\int_{1}^{3} \frac{1}{x + 1} d x

\int_{1}^{3} (x^{2} + 2) d x

What is the integral from 0 to 3 of t/(sqrt(t^2+16)) ?
The integral from 0 to 3 of t/(sqrt(t^2+16)) is 1