Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Exercise 7.1 · Q6: ∫(4e3x+1) dx\int (4e^{3x} + 1)\,dx

Exercise 7.1Properties of indefinite integrals · Section 7.2.1
The question

Find the following integral:

∫(4e3x+1)dx\int\left(4e^{3x} + 1\right)dx
Exam answer

Let I=∫(4e3x+1)dxI = \displaystyle\int\left(4e^{3x} + 1\right)dx⁠.

I=4∫e3x dx+∫1 dx=4⋅e3x3+x+C\begin{aligned} I &= 4\int e^{3x}\,dx + \int 1\,dx \\ &= 4\cdot\frac{e^{3x}}{3} + x + C \end{aligned}

∴ I=43e3x+x+C\therefore\ I = \dfrac43e^{3x} + x + C⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

An integral of a sum is the sum of the integrals, and constant factors come outside. That leaves two standard pieces: ∫e3x dx=13e3x+C\int e^{3x}\,dx = \frac13 e^{3x} + C (the inner 3x3x costs a factor 13\frac13⁠) and ∫1 dx=x+C\int 1\,dx = x + C⁠.

Given, to find, and the plan
Given
The integrand 4e3x+14e^{3x} + 1⁠, defined for every real xx⁠.
To find
∫(4e3x+1) dx\int (4e^{3x} + 1)\,dx⁠, the family of all anti derivatives.
Plan
(1) Split with Properties (III) and (IV). (2) Integrate e3xe^{3x} by inspection. (3) Integrate the constant 11⁠. (4) Assemble with one CC⁠. (5) Differentiate back to check.

Basics you need

1Sum and constant multiple rules for integrals

∫(f+g) dx=∫f dx+∫g dx\int (f + g)\,dx = \int f\,dx + \int g\,dx and ∫kf dx=k∫f dx\int kf\,dx = k\int f\,dx (Properties (III) and (IV), Section 7.2.1). Lesson: integration as anti differentiation

2The exponential

ddxex=ex\frac{d}{dx}e^x = e^x⁠, so ∫ex dx=ex+C\int e^x\,dx = e^x + C (item (xi) of the standard formulae, Section 7.2). With an inner 3x3x the chain rule adds a factor 33⁠. Basics: the chain rule backwards

3Integral of a constant

ddx(x)=1\frac{d}{dx}(x) = 1⁠, so ∫dx=x+C\int dx = x + C (item (i), Section 7.2). Basics: powers of x and the power rule

Solution

  1. Split the integral

    ∫(4e3x+1)dx=4∫e3x dx+∫1 dx.\int\left(4e^{3x} + 1\right)dx = 4\int e^{3x}\,dx + \int 1\,dx.
    Property (III) splits the sum into two integrals, and Property (IV) moves the constant 44 outside. Each piece is now a single standard type we can handle on its own.
    Properties (III) and (IV), Section 7.2.1 (together, Property (V)). Lesson: integration as anti differentiation

  2. Integrate e3xe^{3x} by inspection

    The guess e3xe^{3x} differentiates to 3e3x3e^{3x} (chain rule, inner 3x3x⁠), so divide by 33⁠:

    ddx(13e3x)=e3x,so∫e3x dx=13e3x+C1.\begin{aligned} \frac{d}{dx}\left(\tfrac13 e^{3x}\right) &= e^{3x}, \quad\text{so} \\ \int e^{3x}\,dx &= \tfrac13 e^{3x} + C_1. \end{aligned}
    The exponential reproduces itself; the only change is the factor 33 from the exponent, which we cancel with 13\frac13⁠.
    Item (xi) of the standard formulae; chain rule, Theorem 4, Section 5.3.1 (Chapter 5). Basics: the chain rule backwards

  3. Integrate the constant 1

    ∫1 dx=x+C2.\int 1\,dx = x + C_2.
    We need a function whose slope is 11 everywhere: the line y=xy = x (plus any constant).
    Standard formulae, item (i), Section 7.2: ∫dx=x+C\int dx = x + C⁠. Basics: the power rule

  4. Assemble the answer

    ∫(4e3x+1)dx=4⋅13e3x+x+C=43e3x+x+C.\begin{aligned} \int\left(4e^{3x} + 1\right)dx &= 4\cdot\tfrac13 e^{3x} + x + C \\ &= \tfrac43 e^{3x} + x + C. \end{aligned}
    The constants 4C1+C24C_1 + C_2 combine into one arbitrary constant CC⁠. The 44 multiplies the whole anti derivative 13e3x\frac13 e^{3x}⁠, giving 43e3x\frac43 e^{3x}⁠.
    Note after Example 2: one constant of integration in the final answer. Example 2

  5. Check: differentiate the answer back

    ddx(43e3x+x+C)=43⋅3e3x+1+0=4e3x+1.  ✓\begin{aligned} \frac{d}{dx}\left(\tfrac43 e^{3x} + x + C\right) &= \tfrac43\cdot 3e^{3x} + 1 + 0 \\ &= 4e^{3x} + 1. \;\checkmark \end{aligned}
    The derivative is the integrand for every real xx⁠, so the family is correct on the whole real line.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation
Result ∫(4e3x+1)dx=43e3x+x+C,\int\left(4e^{3x} + 1\right)dx = \tfrac43 e^{3x} + x + C,

valid for every real xx⁠. (Textbook answer: 43e3x+x+C\frac43 e^{3x} + x + C⁠.)

Check it with numbers

At x=0.5x = 0.5 with h=0.001h = 0.001⁠, for F(x)=43e3x+xF(x) = \frac43 e^{3x} + x⁠:

F(0.501)−F(0.499)0.002=6.494539100−6.4566855340.002≈18.92678.\begin{aligned} &\frac{F(0.501) - F(0.499)}{0.002} \\ &= \frac{6.494539100 - 6.456685534}{0.002} \\ &\approx 18.92678. \end{aligned}

The integrand is 4e1.5+1≈4(4.481689)+1=18.926764e^{1.5} + 1 \approx 4(4.481689) + 1 = 18.92676⁠. They agree to six significant figures (the small gap is due to the step hh⁠).

Common mistakes
  1. Writing 12e3x12e^{3x}⁠. That is the derivative of 4e3x4e^{3x}⁠. Integrating divides by 33⁠; differentiating multiplies by 33⁠.
  2. Dropping the xx⁠. The constant 11 in the integrand integrates to xx⁠, not to 00 and not to 11⁠.
  3. Forgetting + C+\,C⁠. The question asks for the integral, which is the whole family.
What to take away
  1. Split, pull out constants, integrate each standard piece, then collect the constants into one CC⁠.
  2. ∫eax dx=1aeax+C\int e^{ax}\,dx = \frac1a e^{ax} + C for a≠0a \ne 0⁠; ∫k dx=kx+C\int k\,dx = kx + C⁠.

Every fact used, and where it comes from

FactUsed inSource
∫(f+g)=∫f+∫g\int (f+g) = \int f + \int g⁠; ∫kf=k∫f\int kf = k\int fStep 1Properties (III), (IV), Section 7.2.1
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 2Standard formulae, item (xi), Section 7.2
Chain ruleSteps 2, 5Theorem 4, Section 5.3.1 (Chapter 5)
∫dx=x+C\int dx = x + CStep 3Standard formulae, item (i), Section 7.2
One constant of integrationStep 4Note after Example 2, Section 7.2.1
Differentiation and integration undo each otherStep 5Property (I), Section 7.2.1
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →