Calculus Examples

$\int_{0}^{x} \sqrt{t^{2} + t^{4}} d t$

Step 1

Simplify terms.

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Factor $t^{2}$ out of $t^{2} + t^{4}$ .

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Multiply by $1$ .

$\frac{d}{d x} [\int_{0}^{x} \sqrt{t^{2} \cdot 1 + t^{4}} d t]$

Factor $t^{2}$ out of $t^{4}$ .

$\frac{d}{d x} [\int_{0}^{x} \sqrt{t^{2} \cdot 1 + t^{2} t^{2}} d t]$

Factor $t^{2}$ out of $t^{2} \cdot 1 + t^{2} t^{2}$ .

$\frac{d}{d x} [\int_{0}^{x} \sqrt{t^{2} (1 + t^{2})} d t]$

Rewrite $1$ as $1^{2}$ .

$\frac{d}{d x} [\int_{0}^{x} \sqrt{t^{2} (1^{2} + t^{2})} d t]$

Pull terms out from under the radical.

$\frac{d}{d x} [\int_{0}^{x} t \sqrt{1^{2} + t^{2}} d t]$

One to any power is one.

$\frac{d}{d x} [\int_{0}^{x} t \sqrt{1 + t^{2}} d t]$

Step 2

Take the derivative of $\int_{0}^{x} t \sqrt{1 + t^{2}} d t$ with respect to $x$ using Fundamental Theorem of Calculus.

$x \sqrt{1 + x^{2}}$