Solve ∫ (from x to 7) of (3x+5/2t^2-16/2)dt

$$\int _ { x } ^ { 7 } ( 3 x + \frac { 5 } { 2 } t ^ { 2 } - \frac { 16 } { 2 } ) d t$$

Evaluate

$$-\frac{\left(x-7\right)\left(5x^{2}+53x+197\right)}{6}$$

Solution Steps

Divide 16 by 2 to get 8.

$$\int _{x}^{7}3x+\frac{5}{2}t^{2}-8\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 3x+\frac{5t^{2}}{2}-8\mathrm{d}t$$

Integrate the sum term by term.

$$\int 3x\mathrm{d}t+\int \frac{5t^{2}}{2}\mathrm{d}t+\int -8\mathrm{d}t$$

Factor out the constant in each of the terms.

$$3\int x\mathrm{d}t+\frac{5\int t^{2}\mathrm{d}t}{2}+\int -8\mathrm{d}t$$

Find the integral of x using the table of common integrals rule \int a\mathrm{d}t=at.

$$3xt+\frac{5\int t^{2}\mathrm{d}t}{2}+\int -8\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply \frac{5}{2} times \frac{t^{3}}{3}.

$$3xt+\frac{5t^{3}}{6}+\int -8\mathrm{d}t$$

Find the integral of -8 using the table of common integrals rule \int a\mathrm{d}t=at.

$$3xt+\frac{5t^{3}}{6}-8t$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$3x\times 7+\frac{5}{6}\times 7^{3}-8\times 7-\left(3xx+\frac{5}{6}x^{3}-8x\right)$$

Simplify.

$$-\frac{\left(-7+x\right)\left(197+53x+5x^{2}\right)}{6}$$

Differentiate w.r.t. x

$$-\frac{5x^{2}}{2}-6x+29$$