8.1 The Fundamental Theorem of Calculus and standard integrals — Practise with help

Work each question closed book first. Check every indefinite integral by differentiating your answer.

Warm-up: remember

Differentiate $y = x^5$ .
Differentiate $y = e^{4x}$ .
Differentiate $y = \sin 3x$ .

Answers

$\dfrac{dy}{dx} = 5x^4$ .
$\dfrac{dy}{dx} = 4e^{4x}$ .
$\dfrac{dy}{dx} = 3\cos 3x$ .

If question 3 was hard, revise the standard derivatives in Section 7.3.

Practise with help

1. Find $\displaystyle\int x^4dx$ .

Hint

Use the power rule for integration. The new power is one more than $4$ .

Next step

Write $\dfrac{x^5}{5}$ , then add the constant of integration.

Full working

\int x^4dx = \frac{x^5}{5} + C.

Check by differentiating: the derivative of $\dfrac{x^5}{5}$ is $x^4$ .

2. Find $\displaystyle\int \left(3x^2 + 2x\right)dx$ .

Hint

Integrate each term separately. The constant of integration is added once, at the end.

Next step

The first term gives $3 \times \dfrac{x^3}{3} = x^3$ , and the second gives $2 \times \dfrac{x^2}{2} = x^2$ .

Full working

\int \left(3x^2 + 2x\right)dx = x^3 + x^2 + C.

Check by differentiating: $3x^2 + 2x$ , which matches the original expression.

3. Find $\displaystyle\int \frac{1}{x^3}dx$ .

Hint

Write the fraction as a power of $x$ before you integrate.

Next step

$\dfrac{1}{x^3} = x^{-3}$ . Raise the power to $-2$ and divide by $-2$ .

Full working

\int x^{-3}dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C.

Leaving the answer as $x^{-3}$ integrated without changing the sign gives a function whose derivative is wrong. Check the sign by differentiating.

4. Find $\displaystyle\int \left(e^{5x} - \sin 2x\right)dx$ .

Hint

Each term needs a factor of $\dfrac{1}{k}$ . Here $k = 5$ for the exponential and $k = 2$ for the sine.

Next step

The exponential integrates to $\dfrac{1}{5}e^{5x}$ . The sine integrates to $-\dfrac{1}{2}\cos 2x$ , keeping the minus sign from the original expression.

Full working

\int \left(e^{5x} - \sin 2x\right)dx = \frac{1}{5}e^{5x} + \frac{1}{2}\cos 2x + C.

The integral of $-\sin 2x$ is $+\dfrac{1}{2}\cos 2x$ . Check the sign by differentiating the answer.

5. Find $\displaystyle\int \left(6\sqrt{x} + \frac{2}{x}\right)dx$ .

Hint

Write $\sqrt{x}$ as $x^{1/2}$ . The second term matches the standard integral of $\dfrac{1}{x}$ .

Next step

$6x^{1/2}$ integrates to $6 \times \dfrac{x^{3/2}}{3/2} = 6 \times \dfrac{2}{3}x^{3/2}$ , which simplifies to $4x^{3/2}$ .

Full working

\int \left(6x^{1/2} + \frac{2}{x}\right)dx = 4x^{3/2} + 2\ln x + C.

The constant multiple $2$ stays with the logarithm. You can also write $4x\sqrt{x}$ for the first term.

6. The gradient of a curve is $\dfrac{dy}{dx} = 4x - 3$ , and the curve passes through the point $(2, 5)$ . Find the equation of the curve.

Hint

Integrate to get $y$ with a constant $C$ , then substitute the coordinates of the given point.

Next step

$y = 2x^2 - 3x + C$ . Substitute $x = 2$ and $y = 5$ to find $C$ .

Full working

y = \int (4x - 3)dx = 2x^2 - 3x + C.

At the point $(2, 5)$ :

5 = 2(2)^2 - 3(2) + C = 8 - 6 + C = 2 + C, \qquad C = 3.

The curve is $y = 2x^2 - 3x + 3$ .