Calculus Examples

$\int_{0}^{1} (x^{16} + 16^{x}) d x$

Step 1

Remove parentheses.

$\int_{0}^{1} x^{16} + 16^{x} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{1} x^{16} d x + \int_{0}^{1} 16^{x} d x$

Step 3

By the Power Rule, the integral of $x^{16}$ with respect to $x$ is $\frac{1}{17} x^{17}$ .

$\frac{1}{17} x^{17}]_{0}^{1} + \int_{0}^{1} 16^{x} d x$

Step 4

The integral of $16^{x}$ with respect to $x$ is $\frac{16^{x}}{\ln (16)}$ .

$\frac{1}{17} x^{17}]_{0}^{1} + \frac{16^{x}}{\ln (16)}]_{0}^{1}$

Step 5

Simplify the answer.

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Step 5.1

Combine $\frac{1}{17} x^{17}]_{0}^{1}$ and $\frac{16^{x}}{\ln (16)}]_{0}^{1}$ .

$\frac{1}{17} x^{17} + \frac{16^{x}}{\ln (16)}]_{0}^{1}$

Step 5.2

Substitute and simplify.

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Step 5.2.1

Evaluate $\frac{1}{17} x^{17} + \frac{16^{x}}{\ln (16)}$ at $1$ and at $0$ .

$(\frac{1}{17} \cdot 1^{17} + \frac{16^{1}}{\ln (16)}) - (\frac{1}{17} \cdot 0^{17} + \frac{16^{0}}{\ln (16)})$

Step 5.2.2

Simplify.

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Step 5.2.2.1

One to any power is one.

$\frac{1}{17} \cdot 1 + \frac{16^{1}}{\ln (16)} - (\frac{1}{17} \cdot 0^{17} + \frac{16^{0}}{\ln (16)})$

Step 5.2.2.2

Multiply $\frac{1}{17}$ by $1$ .

$\frac{1}{17} + \frac{16^{1}}{\ln (16)} - (\frac{1}{17} \cdot 0^{17} + \frac{16^{0}}{\ln (16)})$

Step 5.2.2.3

Evaluate the exponent.

$\frac{1}{17} + \frac{16}{\ln (16)} - (\frac{1}{17} \cdot 0^{17} + \frac{16^{0}}{\ln (16)})$

Step 5.2.2.4

Raising $0$ to any positive power yields $0$ .

$\frac{1}{17} + \frac{16}{\ln (16)} - (\frac{1}{17} \cdot 0 + \frac{16^{0}}{\ln (16)})$

Step 5.2.2.5

Multiply $\frac{1}{17}$ by $0$ .

$\frac{1}{17} + \frac{16}{\ln (16)} - (0 + \frac{16^{0}}{\ln (16)})$

Step 5.2.2.6

Anything raised to $0$ is $1$ .

$\frac{1}{17} + \frac{16}{\ln (16)} - (0 + \frac{1}{\ln (16)})$

Step 5.2.2.7

Add $0$ and $\frac{1}{\ln (16)}$ .

$\frac{1}{17} + \frac{16}{\ln (16)} - \frac{1}{\ln (16)}$

Step 5.2.2.8

Combine the numerators over the common denominator.

$\frac{1}{17} + \frac{16 - 1}{\ln (16)}$

Step 5.2.2.9

Subtract $1$ from $16$ .

$\frac{1}{17} + \frac{15}{\ln (16)}$

Step 6

Simplify the denominator.

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Step 6.1

Rewrite $\ln (16)$ as $\ln (2^{4})$ .

$\frac{1}{17} + \frac{15}{\ln (2^{4})}$

Step 6.2

Expand $\ln (2^{4})$ by moving $4$ outside the logarithm.

$\frac{1}{17} + \frac{15}{4 \ln (2)}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{17} + \frac{15}{4 \ln (2)}$

Decimal Form:

$5.46892993 \dots$

Step 8